Continuity And Differentiability

The given function is f(x) = 5x -3

At x = 0, f(0) = 5 × 0 – 3 = -3

Thus,

Therefore, f is continuous at x = 0

At x = -3, f(-3) = 5 × (-3) – 3 = -18

Thus,

Therefore, f is continuous at x = -3

At x = 5, f(5) = 5 × 5 – 3 = 22

= 5 × 5 – 3 = 22

Thus,

Therefore, f is continuous at x = 5

The given function is f(x) = 2x² – 1

At x = 3, f(x) = f(3) = 2 × 3² – 1 = 17

As, LHL= RHL = f(3)
Therefore, f is continuous at x = 3

a) The given function is f(x) = x – 5

We know that f is defined at every real number k and its value at k is k – 5.

We can see that = k – 5 = f(k)

Thus,

Therefore, f is continuous at every real number and thus, it is continuous function.

The given function is

For any real number k ≠ 5, we get,

Also,

Thus,

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

The given function is

For any real number k ≠ 5, we get,

Also,

Thus,

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.ee, k < 5, or k = 5 or k >5

Now, Case I: k<5

Then, f(k) = 5 – k

= 5 – k = f(k)

Thus,

Hence, f is continuous at all real number less than 5.

Case II: k = 5

Then, f(k) = f(5) = 5 – 5 = 0

= 5 – 5 = 0

= 5 – 5 = 0

Hence, f is continuous at x = 5.

Case III: k > 5

Then, f(k) = k – 5

= k – 5 = f(k)

Thus,

Hence, f is continuous at all real number greater than 5.

Therefore, f is a continuous function.

It is given that function f (x) = xⁿ

We can see that f is defined at all positive integers, n and the value of f at n is nⁿ.

= nⁿ

Thus,

Therefore, f is continuous at x =n, where n is a positive integer.

It is given that

Case I: x = 0

We can see that f is defined at 0 and its value at 0 is 0.

LHL = RHL = f(0)
Hence, f is continuous at x = 0.

Case II: x = 1

We can see that f is defined at 1 and its value at 1 is 1.

= 1

= 5

Hence, f is not continuous at x = 1.

Case III: x = 2

We can see that f is defined at 2 and its value at 2 is 5
LHL:

here f(2 - h) = 5, as h → 0 ⇒ 2 - h → 2

RHL:

LHL = RHL = f(2)

here f(2 + h) = 5, as h → 0 ⇒ 2 + h → 2
Hence, f is continuous at x = 2.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k >2

Now, Case I: k < 2

Then, f(k) = 2k + 3

= 2k + 3= f(k)

Thus,

Hence, f is continuous at all real number less than 2.

Case II: k = 2

= 2×2 + 3 = 7

= 2×2 - 3 = 1

Hence, f is not continuous at x = 2.

Case III: k > 2

Then, f(k) = 2k - 3

= 2k – 3 = f(k)

Thus,

Hence, f is continuous at all real number greater than 2.

Therefore, x = 2 is the only point of discontinuity of f.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 5 cases i.e., k < -3, k = -3, -3 < k < 3, k = 3 or k > 3

Now, Case I: k < -3

Then, f(k) = -k + 3

= -k + 3= f(k)

Thus,

Hence, f is continuous at all real number x < -3.

Case II: k = -3

f(-3) = -(-3) + 3 = 6

=-(-3) + 3 = 6

= -2×(-3) = 6

Hence, f is continuous at x = -3.

Case III: -3 < k < 3

Then, f(k) = -2k

= -2k = f(k)

Thus,

Hence, f is continuous in (-3,3).

Case IV: k = 3

= -2×(3) = -6

= 6 × 3 + 2 = 20

Hence, f is not continuous at x = 3.

Case V: k > 3

Then, f(k) = 6k + 2

= 6k + 2= f(k)

Thus,

Hence, f is continuous at all real number x < 3.

Therefore, x = 3 is the only point of discontinuity of f.

The given function is

We know that if x > 0

⇒ |x| = -x and

x > 0

⇒ |x| = x

So, we can rewrite the given function as:

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 0, or k = 0 or k >0.

Now, Case I: k < 0

Then, f(k) = -1

= -1= f(k)

Thus,

Hence, f is continuous at all real number less than 0.

Case II: k = 0

= -1

= 1

Hence, f is not continuous at x = 0.

Case III: k > 0

Then, f(k) = 1

= 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 0 is the only point of discontinuity of f.

The given function is

We know that if x < 0

⇒ |x| = -x

So, we can rewrite the given function as:

⇒ f(x) = -1 for all x ϵ R

Let k be the point on a real line.

Then, f(k) = -1

= -1= f(k)

Thus,

Therefore, the given function is a continuous function.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k >1

Now, Case I: k < 1

Then, f(k) = k² + 1

= k² + 1= f(k)

Thus,

Hence, f is continuous at all real number less than 1.

Case II: k = 1

Then, f(k) = f(1) = 1 + 1 = 2

= 1² + 1 = 2

= 1 + 1 = 2

Hence, f is continuous at x = 1.

Case III: k > 1

Then, f(k) = k + 1

= k + 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k > 2

Now, Case I: k < 2

Then, f(k) = k³ - 3

= k³ - 3= f(k)

Thus,

Hence, f is continuous at all real number less than 2.

Case II: k = 2

Then, f(k) = f(2) = 2³ - 3 = 5

= 2³ - 3 = 5

= 2² + 1 = 5

Hence, f is continuous at x = 2.

Case III: k > 2

Then, f(k) = 2² + 1 = 5

= 2² + 1 = 5 = f(k)

Thus,

Hence, f is continuous at all real number greater than 2.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

Case I: k < 1

Then, f(k) = k¹⁰ - 1

= k¹⁰ - 1= f(k)

Thus,

Hence, f is continuous at all real number less than 1.

Case II: k = 1

Then, f(k) = f(1) = 1¹⁰ - 1 = 0

= 1¹⁰ - 1 = 0

= 1² = 1

Hence, f is not continuous at x = 1.

Case III: k > 1

Then, f(k) = 1² = 1

= 1² = 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

Case I: k < 1

Then, f(k) = k + 5

= k + 5 = f(k)

Thus,

Hence, f is continuous at all real number less than 1.

Case II: k = 1

Then, f(k) = f(1) = 1 + 5 = 6

= 1 + 5 = 6

= 1 - 5 = -4

Hence, f is not continuous at x = 1.

Case III: k > 1

Then, f(k) = k -5

= k - 5

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

The given function is

The function f is defined at all points of the interval [0,10].

Let k be the point in the interval [0,10].

Then, we have 5 cases i.e., 0≤ k < 1, k = 1, 1 < k < 3, k = 3 or 3 < k ≤ 10.

Now, Case I: 0≤ k < 1

Then, f(k) = 3

= 3= f(k)

Thus,

Hence, f is continuous in the interval [0,10).

Case II: k = 1

f(1) = 3

= 3

= 4

Hence, f is not continuous at x = 1.

Case III: 1 < k < 3

Then, f(k) = 4

= 4 = f(k)

Thus,

Hence, f is continuous in (1, 3).

Case IV: k = 3

= 4

= 5

Hence, f is not continuous at x = 3.

Case V: 3 < k ≤ 10

Then, f(k) = 5

= 5 = f(k)

Thus,

Hence, f is continuous at all points of the interval (3, 10].

Therefore, x = 1 and 3 are the points of discontinuity of f.

The given function is

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1 or k < 1.

Now, Case I: k < 0

Then, f(k) = 2k

= 2k= f(k)

Thus,

Hence, f is continuous at all points x, s.t. x < 0.

Case II: k = 0

f(0) = 0

= 2 × 0 = 0

= 0

Hence, f is continuous at x = 0.

Case III: 0 < k < 1

Then, f(k) = 0

= 0 = f(k)

Thus,

Hence, f is continuous in (0, 1).

Case IV: k = 1

Then f(k) = f(1) = 0

= 0

= 4 × 1 = 4

Hence, f is not continuous at x = 1.

Case V: k < 1

Then, f(k) = 4k

= 4k = f(k)

Thus,

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, x = 1 is the only point of discontinuity of f.

The given function is

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < -1, k = -1, -1 < k < 1, k = 1 or k > 1.

Now, Case I: k < 0

Then, f(k) = -2

= -2= f(k)

Thus,

Hence, f is continuous at all points x, s.t. x < -1.

Case II: k = -1

f(k) = f(=1) = -2

= -2

= 2 × (-1) = -2

Hence, f is continuous at x = -1.

Case III: -1 < k < 1

Then, f(k) = 2k

= 2k = f(k)

Thus,

Hence, f is continuous in (-1, 1).

Case IV: k = 1

Then f(k) = f(1) = 2 × 1 = 2

= 2 × 1 = 2

= 2

Hence, f is continuous at x = 1.

Case V: k > 1

Then, f(k) = 2

= 2 = f(k)

Thus,

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, f is continuous at all points of the real line.

It is given function is

It is given that f is continuous at x = 3, then, we get,

………………….(1)

And

= 3a + 1

= 3b + 1

f(3) = 3a + 1

Thus, from (1), we get,

3a + 1 = 3b + 3 = 3a + 1

⇒ 3a +1 = 3b + 1

⇒ 3a = 3b + 2

⇒ a = b +

Therefore, the required the relation is a = b + .

It is given that

It is given that f is continuous at x = 0, then, we get,

And

= 0

= 1

Thus, there is no value of for which f is continuous at x = 0

f(1) = 4x + 1 = 4 × 1 + 1 = 5

= 4 × 1 + 1 = 5

Then,

Hence, for any values of, f is continuous at x = 1

It is given that g(x) = x – [x]

We know that g is defined at all integral points.

Let k be ant integer.

Then,

g(k) = k – [-k] = k + k = 2k

And

Therefore, g is discontinuous at all integral points.

It is given that f (x) = x² – sin x + 5

We know that f is defined at x = π

So, at x = π,

f(x) = f(π) = π² -sin π + 5 = π² – 0 + 5 = π² + 5

Now,

Let put x = π + h

If

Thus,

Therefore, the function f is continuous at x = π.

We known that g and k are two continuous functions, then,

g + k, g – k and g.k are also continuous.

First we have to prove that g(x) = sinx and k(x) = cosx are continuous functions.

Now, let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if

g(h) = sinh

= sinhcos0 + coshsin0

= sinh + 0

= sinh

Thus,

Therefore, g is a continuous function…………(1)

Now, let k(x) = cosx

We know that k(x) = cosx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if

Now k(h) = cosh

= coshcos0 - sinhsin0

= cosh - 0

= cosh

Thus,

Therefore, k is a continuous function……………….(2)

So, from (1) and (2), we get,

(a) f(x) = g(x) + k(x) = sinx + cosx is a continuous function.

(b) f(x) = g(x) - k(x) = sinx - cosx is a continuous function.

(c) f(x) = g(x) × k(x) = sinx × cosx is a continuous function.

We know that if g and h are two continuous functions, then,

(i)

(ii)

(iii)

So, first we have to prove that g(x) = sinx and h(x) = cosx are continuous functions.

Let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = k + h

So, if

g(k) = sink

= sinkcos0 + cosksin0

= sink + 0

= sink

Thus,

Therefore, g is a continuous function…………(1)

Let h(x) = cosx

We know that h(x) = cosx is defined for every real number.

Let k be a real number. Now, put x = k + h

So, if

h(k) = sink

= coskcos0 - sinksin0

= cosk - 0

= cosk

Thus,

Therefore, g is a continuous function…………(2)

So, from (1) and (2), we get,

Thus, cosecant is continuous except at x = np, (n ϵ Z)

Thus, secant is continuous except at x = , (n ϵ Z)

Thus, cotangent is continuous except at x = np, (n ϵ Z)

It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k < 0,

Then f(k) =

Thus, f is continuous at all points x that is x < 0.

Case II: k > 0,

Then f(k) = c + 1

Thus, f is continuous at all points x that is x > 0.

Case III: k = 0

Then f(k) = f(0) = 0 + 1 = 1

= 1

= 1

Hence, f is continuous at x = 0.

Therefore, f is continuous at all points of the real line.

It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) =

Thus, f is continuous at all points x that is x ≠ 0.

Case II: k = 0

Then f(k) = f(0) = 0

We know that -1 ≤ ≤ 1, x ≠ 0

⇒ x² ≤ ≤ 0

⇒

⇒

Similarly,

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) = sink - cosk

Thus, f is continuous at all points x that is x ≠ 0.

Case II: k = 0

Then f(k) = f(0) = 0

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

It is given that

Also, it is given that function f is continuous at x =,

So, if f is defined at x = and if the value of the f at x = equals the limit of f at x = .

We can see that f is defined at x = and f = 3

Now, let put x =

Then,

⇒

⇒

Therefore, the value of k is 6.

It is given that

Also, it is given that function f is continuous at x = 2,

So, if f is defined at x = 2 and if the value of the f at x = 2 equals the limit of f at x = 2.

We can see that f is defined at x = 2 and

f(2) = k(2)² = 4k

⇒

⇒ k × 2² = 3 = 4k

⇒ 4k = 3 = 4k

⇒ 4k = 3

⇒ k =

Therefore, the required value of k is .

It is given that

Also, it is given that function f is continuous at x = k,

So, if f is defined at x = p and if the value of the f at x = k equals the limit of f at x = k.

We can see that f is defined at x = p and

f(π) = kπ + 1

⇒

⇒ kπ + 1 = cosπ = kπ + 1

⇒ kπ + 1 = -1 = kπ + 1

⇒ k =

Therefore, the required value of k is.

It is given that

Also, it is given that function f is continuous at x = 5,

So, if f is defined at x = 5 and if the value of the f at x = 5 equals the limit of f at x = 5.

We can see that f is defined at x = 5 and

f(5) = kx + 1 = 5k + 1

⇒

⇒ 5k + 1 = 15 -5 = 5k + 1

⇒ 5k + 1 = 10

⇒ 5k = 9

⇒ k =

Therefore, the required value of k is.

It is given function is

We know that the given function f is defined at all points of the real line.

Thus, f is continuous at x = 2, we get,

⇒

⇒ 5 = 2a + b = 5

⇒ 2a + b = 5………………(1)

Thus, f is continuous at x = 10, we get,

⇒

⇒ 10a + b = 21 =21

⇒ 10a + b = 21………………(2)

On subtracting eq. (1) from eq. (2), we get,

8a = 16

⇒ a = 2

Thus, putting a = 2 in eq. (1), we get,

2 × 2 + b = 5

⇒ 4 + b = 5

⇒ b = 1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 resp.

It is given function is f(x) = cos (x²)

This function f is defined for every real number and f can be written as the composition of two function as,

f = goh, where, g(x) = cosx and h(x) = x²

First we have to prove that g(x) = cosx and h(x) = x² are continuous functions.

We know that g is defined for every real number.

Let k be a real number.

Then, g(k) =cos k

Now, put x = k + h

If

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, g(x) = cosx is continuous function.

Now, h(x) = x²

So, h is defined for every real number.

Let c be a real number, then h(c) = c²

Therefore, h is a continuous function.

We know that for real valued functions g and h,

Such that (fog) is continuous at c.

Therefore, f(x) = (goh)(x) = cos(x²) is a continuous function.

It is given that f(x) = |cosx|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = cosx

First we have to prove that g(x) = |x| and h(x) = cosx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = cosx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(cosx) = |cosx| is a continuous function.

It is given that f(x) = sin|x|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = sinx

First we have to prove that g(x) = |x| and h(x) = sinx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = sinx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If

= sinkcos0 + cosksin0

= sink

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(sinx) = |sinx| is a continuous function.

It is given that f(x) = |x| - |x + 1|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = |x + 1|

Then, f = g - h

First we have to prove that g(x) = |x| and h(x) = |x + 1| are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

g(x) = |x + 1| can be written as

Now, h is defined for all real number.

Let k be a real number.

Case I: If k < -1,

Then h(k) = -(k + 1)

And

Thus,

Therefore, h is continuous at all points x, i.e., x < -1

Case II: If k > -1,

Then h(k) = k + 1 and

Thus,

Therefore, h is continuous at all points x, i.e., x > -1.

Case III: If k = -1,

Then, h(k) = h(-1) = -1 + 1 = 0

Therefore, g is continuous at x = -1

From the above 3 cases, we get that h is continuous at all points.

Hence, g and h are continuous function.

Therefore, f = g – h is also a continuous function.

Given: sin(x² + 5)

Let y = sin(x² + 5)

= cos(x² + 5).(2x + 0)

= cos(x² + 5).(2x)

= 2x.cos(x² + 5)

Given: cos(sinx)

Let y = cos(sinx)

= -sin(sinx).cosx

= -cosx.sin(sinx)

Given: sin(ax + b)

Let y = sin(ax + b)

= cos(ax + b).(a + 0)

= cos(ax + b). (a)

= a.cos(ax + b)

Given: sec (tan(√x))

Let y= sec (tan(√x))

Given:

Let

We know that

= a cos (ax + b) sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d)

Given: cos x³ . sin² (x⁵)

Let y = cos x³ . sin² (x⁵)

We know that,

= cosx³.2sin(x⁵).cos(x⁵)(5x⁴) + sin²(x⁵).(-sinx³).(3x²)

= 10x⁴.cosx³.sin(x⁵).cos(x⁵)-(3x²).sin² (x⁵).(sinx³)

Let

Now,

Therefore,

Given: cos√x

Let y = cos√x

Given: f(x)=|x-1|, x ∈R

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left hand limit of function f at x=1

because, {h < 0 ⇒ |h|= -h}

= -1

Now, let we consider the right hand limit of function f at x=1

because, {h>0 ⇒ |h|= h}

= 1

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

Given: f(x) =[x], 0 < x <3

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left-hand limit of function f at x=1

because, {h<0=> |h|= -h}

Let we consider the right hand limit of function f at x=1

= 0

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

Let we consider the left hand limit of function f at x=2

= =

Now, let we consider the right hand limit of function f at x=2

= 0

Because, left hand limit is not equal to right hand limit of function f at x=2, so f is not differentiable at x=2.

It is given that 2x + 3y = sin x

Differentiating both sides w.r.t. x, we get,

It is given that 2x + 3y = sin y

Differentiating both sides w.r.t. x, we get,

It is given that ax + by² = cos y

Differentiating both sides w.r.t. x, we get,

It is given that xy + y² = tan x + y

Differentiating both sides w.r.t. x, we get,

It is given that x² + xy + y² = 100

Differentiating both sides w.r.t. x, we get,

It is given that x³ + x²y + xy² + y³ = 81

Differentiating both sides w.r.t. x, we get,

It is given that sin² y + cos xy = π

Differentiating both sides w.r.t. x, we get,

It is given that sin² x + cos² y = 1

Differentiating both sides w.r.t. x, we get,

It is given that:


Assumption: Let x = tan θ, putting it in y, we get,



we know by the formula that,

Putting this in y, we get,

y = tan^-1(tan3θ)

y = 3(tan^-1x)

Differentiating both sides, we get,

It is given that,

y =

On comparing both sides, we get,

Now, differentiating both sides, we get,

It is given that y =

Now, differentiating both sides, we get,

It is given that y =

Differentiating both sides w.r.t. x, we get,

It is given that y =

Differentiating both sides w.r.t. x, we get,

It is given that y =

Differentiating w.r.t. x, we get,

Let y =

By using the quotient rule, we get

Let y =

Now, by using the chain rule, we get,

Thus,

Let y =

So, by using the chain rule, we get,

Let y = sin(tan^-1 e^x)

So, by using chain rule, we get

Let y = log(cose^x)

So, by using the chain rule, we get,

Let

Let

Then,

Now, differentiating both sides we get,

let y = log(logx)

So, by using chain rule, we get,

Let

So, by using the quotient rule, we get,

Let y = cos(logx + e^x)

So, by using chain rule, we get,

Given: cos x . cos 2x . cos 3x

Let y= cos x . cos 2x . cos 3x

Taking log on both sides, we get

log y = log(cos x . cos 2x . cos 3x)

⇒log y = log(cos x) + log(cos 2x) + log(cos 3x)

Now, differentiate both sides with respect to x

Given:

Let

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Given: (log x)^{cos x}

Let y=(log x)^{cos x}

Taking log on both sides, we get

log y = log(log x)^{cos x}

⇒log y = cos x.log(log x)

Now, differentiate both sides with respect to x

Given: x^x – 2^{sin x}

Let y= x^x – 2^{sin x}

Let y = u - v

⇒ u = x^x and v = 2^{sin x}

For, u = x^x

Taking log on both sides, we get

log u = log x^x

⇒log u = x.log(x)

Now, differentiate both sides with respect to x

For, v = 2^{sin x}

Taking log on both sides, we get

log v = log 2^{sin x}

⇒log v = sin x. log (2)

Now, differentiate both sides with respect to x

Because, y = u - v

Given: (x + 3)². (x + 4)³. (x + 5)⁴

Let y= (x + 3)². (x + 4)³. (x + 5)⁴

Taking log on both sides, we get

log y = log((x + 3)². (x + 4)³. (x + 5)⁴)

⇒log y = log(x + 3)² + log(x + 4)³ + log(x + 5)⁴

⇒log y = 2.log(x + 3) + 3.log(x + 4) + 4.log(x + 5)⁴

Now, differentiate both sides with respect to x

= (x + 3)(x + 4)²(x + 5)³(9x² + 70x + 133)

Given:

Let y=

Also, Let y = u + v

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Because, y = u + v

Given: (log x)^x + x^{log x}

Let y= (log x)^x + x^{log x}

Let y = u + v

⇒ u = (log x)^x and v = x^{log x}

For, u =(log x)^x

Taking log on both sides, we get

log u =log (log x)^x

⇒log u = x.log (log(x))

Now, differentiate both sides with respect to x

For, v = x^{log x}

Taking log on both sides, we get

log v =log( x^{log x)}

⇒log v = log x. log x

Now, differentiate both sides with respect to x

Because, y = u + v

Given:

Let y=

Let y = u + v

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Now, differentiate both sides with respect to x

Because, y = u + v

Given: x^{sin x} + (sin x)^{cos x}

Let y= x^{sin x} + (sin x)^{cos x}

Let y = u + v

⇒ u = x^{sin x} and v = (sin x)^{cos x}

For, u = x^{sin x}

Taking log on both sides, we get

log u =log (x^{sin x} )

⇒log u = sin x.log(x)

Now, differentiate both sides with respect to x

For, v = (sin x)^{cos x}

Taking log on both sides, we get

log v =log(sin x)^{cos x}

⇒log v = cos x. log (sin x)

Now, differentiate both sides with respect to x

Because, y = u + v

Given:

Let y=

Let y = u + v

Taking log on both sides, we get

⇒log u = x.cos x.log x

Now, differentiate both sides with respect to x

Taking log on both sides, we get

⇒ log v = log (x² + 1) – log (x² – 1)

Now, differentiate both sides with respect to x

Because, y = u + v

Given:

Let y=

Let y = u + v

Taking log on both sides, we get

⇒log u = x.log(x cos x)

⇒ log u = x(log x + log (cos x))

⇒ log u = x(log x) +x(log (cos x))

Now, differentiate both sides with respect to x

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Because, y = u + v

Given: x^y + y^x = 1

Let y= x^y + y^x = 1

Let u = x^y and v = y^x

Then, ⇒ u + v = 1

For, u = x^y

Taking log on both sides, we get

Log u =log x^y

⇒log u = y.log(x)

Now, differentiate both sides with respect to x

For, v = y^x

Taking log on both sides, we get

Log v =log y^x

⇒log v = x.log(y)

Now, differentiate both sides with respect to x

Given: y^x = x^y

Taking log on both sides, we get

log y^x =log x^y

⇒x log y = y log x

Now, differentiate both sides with respect to x

Given: (cos x)^y = (cos y)^x

Taking log on both sides, we get

log (cos x)^y =log (cos y)^x

⇒y log (cos x) = x log (cos y)

Now, differentiate both sides with respect to x

Given: xy = e^{(x – y)}

Taking log on both sides, we get

log (xy) = log (e^{(x – y)})

⇒ log x + log y = (x - y) log e

⇒ log x + log y = (x - y) .1

⇒ log x + log y = (x - y)

Now, differentiate both sides with respect to x

Given: f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸)

Taking log on both sides, we get

log f (x) =log (1 + x) + log (1 + x²) + log (1 + x⁴) + log (1 + x⁸)

Now, differentiate both sides with respect to x

Given: (x² – 5x + 8) (x³ + 7x + 9)

Let y=(x² – 5x + 8) (x³ + 7x + 9)

(i) By applying product rule differentiate both sides with respect to x

(ii) by expanding the product to obtain a single polynomial

y = (x² – 5x + 8) (x³ + 7x + 9)

y = x⁵ + 7x³ + 9x² - 5x⁴ – 35x² - 45x + 8x³ + 56x + 72

y = x⁵ - 5x⁴ + 15x³ - 26x² + 11x + 72

Now, differentiate both sides with respect to x

(iii) by logarithmic differentiation

y = (x² – 5x + 8) (x³ + 7x + 9)

Taking log on both sides, we get

log y = log ((x² – 5x + 8) (x³ + 7x + 9))

log y = log (x² – 5x + 8) + log (x³ + 7x + 9)

Now, differentiate both sides with respect to x

From equation (i),(ii)and(iii), we can say that value of given function after differentiating by all the three methods is same.

To prove:

Let y=u.v.w=u.(v.w)

(a) by applying product rule differentiate both sides with respect to x

(b) Taking log on both sides, we get

as, y=u.v.w

log y = log (u.v.w)

log y = log u + log v + log w

Now, differentiate both sides with respect to x

It is given that

x = 2at², y = at⁴

So, now

= 2a.2t

= 4at ………… (1)

And

= a.4.t³

= 4at³………… (2)

Therefore, form equation (1) and (2). we get

Hence, the value of is t²

It is given that

x = a cos θ, y = b cos θ

Then, we have

= a(-sinƟ)

= -asinƟ………… (1)

= b(-sinƟ)

= -bsinƟ …… (2)

From equation (1) and (2), we get

Hence, the value of is

It is given that

x = sin t, y = cos 2t

Then, we have

= cost ………… (1)

= -2sin2t………… (2)

So, equation (1) and (2), we get

, Since sin2t = 2sintcost

= -4sint

Hence, the value of is -4sint

It is given that

x = 4t, y =

Then, we have

= 4 ……… (1)

………… (2)

Therefore, from equation (1) and (2), we get

Hence, the value of is

It is given that

x = cos θ – cos 2θ, y = sin θ – sin 2θ

Then, we have

= -sinθ – (-2sin2θ)

= 2sin2θ - sinθ …………… (1)

= cosθ – 2cos2θ

= -bsinƟ ………… (2)

From equation (1) and (2), we get,

Hence, the value of is

It is given that

x = a (θ – sin θ), y = a (1 + cos θ)

Then, we have

= a(1-cosθ) …………………… (1)

= a[0 + (-sinθ)]

= -asinƟ …………………… (2)

From equation (1) and (2), we get

Hence, the value of is

It is given that

Then, we have

…………………… (1)

…………………… (2)

Therefore, from equation (1) and (2), we get

, [since, cos2t= (2cos²t-1) and also cos2t= (1-2sin²t)]

, [Since, cos3t = 4cos³t-3cost, sin3t = 3sint – 4sin³t]

= -cot3t

Hence, the value of is -cot3t

It is given that

Then, we have

…………………… (1)

= acost…………………… (2)

From equation (1) and (2), we get

= tant

Hence, the value of is tant

It is given that

x = a sec θ, y = b tan θ

Then, we have

= asecθtanθ…………………… (1)

= bsec²θ…………………… (2)

From equation (1) and (2), we get,

Hence, the value of is

It is given that

x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)

Then, we have

= a[-sinθ + θcosθ + sinθ]

= aθcosθ…………………… (1)

= a[cosθ + θsinθ - cosθ]

= aθsinθ…………………… (2)

From (1) and (2) we get,

= tanθ

Hence, the value of is tanθ

It is given that

Now,

Similarly,

Let us consider,

Taking Log on both sides, we get

Therefore,

……………………………..(1)

Now, Consider

Taking Log on both sides, we get

Therefore,

……………..(2)

So, from equation (1) and (2), we get

Therefore, L.H.S. = R.H.S.

Hence Proved

Let us take y= x² + 3x + 2

Now,

= 2x + 3

Therefore,

= 2 + 0

= 2

Let us take y= x²⁰

Now,

= 20x¹⁹

Therefore,

= 20 × 19 × x¹⁸

= 380 x¹⁸

: Let us take y= x . cos x

Now,

= cosx.1 + x(-sinx)

= cosx -xsinx

Therefore,

= -sinx –(sinx + xcosx)

= - (xcosx + 2sinx)

Let us take y = logx

Now,

Therefore,

Let us take y = x³ log x

Now,

= logx.3x² + x³.1/x

= logx .3x² + x²

= x²(1 + 3logx)

Therefore,

= 2x + 6xlogx + 3x

= 5x + 6xlogx

= x(5 + 6logx)

Let us take y = e^x sin 5x

Now,

= e^xsin5x+e^xcos5x.5

= e^x (sin5x + 5cos5x)

= e^x (sin5x + 5cos5x) + e^x (5cos5x – 25sin5x)

= e^x (10cos5x – 24sin5x)

= 2e^x (5cos5x – 12sin5x)

Let us take y = e^6x cos 3x

Now,

= 6e^6xcos3x - 3e^6xsin3x

= 36e^6xcos3x – 18e^6xsin3x -3[sin3x.e^6x.6 + e^6x.cos3x.3]

= 36e^6xcos3x – 18e^6xsin3x – 18e^6xsin3x -9e^6xcos3x

= 27e^6xcos3x -36e^6xsin3x

= 9e^6x(3cos3x – 4sin3x)

Let us take y = tan^–1 x Now,

Let us take y = log (log x)

Now,

= (xlogx)^-1

Let us take y = sin (log x)

Now,

Then,

It is given that y = 5 cos x – 3 sin x

Now, on differentiating we get,

= 5(-sinx) – 3(cosx)

= -(5sinx + cosx)

Then,

= - [5cosx +3(-sinx)]

= -[5cosx-3sinx]

= -y

Therefore,

Hence Proved.

It is given that y = cos^–1 x

Now,

Therefore,

………………………(1)

Now it is given that y = cos^–1 x

⇒ x= cosy

Now putting the value of x in equation (1), we get

It is given that y = 3 cos (log x) + 4 sin (log x)

Now, on differentiating we get,

Again differentiating we get,

Therefore,

x² y₂ + xy₁ + y

= -sin(logx) – 7cos(logx) + 4cos(logx) – 3sin(logx) + 3cos(logx) + 4sin(logx)

= 0

So, x² y₂ + xy₁ + y = 0

Hence Proved

According to given equation, we have,

y = Ae^mx + Be^nx

Then,

= Ame^mx + Bne^nx

Now, on again differentiating we get,

= Am²e^mx + Bn²e^nx

= Am²e^mx + Bn²e^nx – (m+n)( Ame^mx + Bne^nx) + mn(Ae^mx + Be^nx)

= Am²e^mx + Bn²e^nx- Am²e^mx - Bmne^nx -Amne^mx - Bn²e^nx + Amne^mx + Bmne^nx

= 0

Hence Proved

According to given equation, we have,

y = 500e^7x + 600e^–7x

= 3500e^7x - 4200e^-7x

Now, on again differentiating we get,

= 7×3500.e^7x + 7×4200.e^-7x

= 49×500e^7x + 49×600e^-7x

= 49(500e^7x + 600e^-7x)

= 49y

Hence Proved

It is given that

e^y (x + 1) = 1

Now, taking logarithm on both the sides we get,

On differentiating both sides we get,

Again, on differentiating we get,

Hence Proved

: It is given that

y = (tan^–1 x)²

On differentiating we get,

Again differentiating, we get,

So, (1+x²)²y₂ + 2x(1+x²)y₁ = 2

where,

Hence Proved

The given function is f (x) = x² + 2x – 8 and x ∈ [-4, 2].

By Rolle’s Theorem, for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then there exists some c in (a, b) such that f '(c) = 0.

As f(x) = x² + 2x – 8 is a polynomial function,

(a) f(x) is continuous in [-4, 2]

(b) f'(x) = 2x + 2

So, f(x) is differentiable in (-4, 2).

(c) f(a) = f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 =16 – 16 = 0

f(b) = f(2) = (2)² + 2(2) – 8 = 4 + 4 – 8 = 8 – 8 = 0

Hence, f(a) = f(b).

∴ There is a point c ∈ (-4, 2) where f'(c) = 0.

f(x) = x² +2x – 8

f'(x) = 2x + 2

f'(c) = 0

⇒ f'(c) = 2c + 2 = 0

⇒2c = -2

⇒ c = -2 /2

⇒ c = -1 where c = -1 ∈ (-4, 2)

Hence, Rolle’s Theorem is verified.

By Rolle’s Theorem, for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then there exists some c in (a, b) such that f '(c) = 0.

If a function does not satisfy any of the above conditions, then Rolle’s Theorem is not applicable.

(i) f (x) = [x] for x ∈ [5, 9]

As the given function is a greatest integer function,

(a) f(x) is not continuous in [5, 9]

(b) Let y be an integer such that y ∈ (5, 9)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x = y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in [5, 9]

(c) f(a)= f(5) = [5] = 5

f(b) = f(9) = [9] = 9

f(a) ≠ f(b)

Here, f(x) does not satisfy the conditions of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = [x] for x∈[5, 9].

(ii) f (x) = [x] for x ∈ [– 2, 2]

As the given function is a greatest integer function,

(a) f(x) is not continuous in [-2, 2]

(b) Let y be an integer such that y ∈ (-2, 2)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x = y are not equal, f(x) is not differentiable at x = y.

So, f(x) is not differentiable in (-2, 2)

(c) f(a)= f(-2) = [-2] = -2

f(b) = f(2) = [2] = 2

f(a) ≠ f(b)

Here, f(x) does not satisfy the conditions of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = [x] for x∈[-2, 2].

(iii) f (x) = x² – 1 for x ∈ [1, 2]

As the given function is a polynomial function,

(a) f(x) is continuous in [1, 2]

(b) f'(x) = 2x

So, f(x) is differentiable in [1, 2]

(c) f(a) = f(1) = 1² – 1 = 1 – 1 = 0

f(b) = f(2) = 2² - 1 = 4 – 1 = 3

f(a) ≠ f(b)

Here, f(x) does not satisfy a condition of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = x² – 1 for x∈[1, 2].

Given: f : [-5, 5] → R is a differentiable function.

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

We know that a differentiable function is a continuous function.

So,

(a) f is continuous on [-5, 5]

(b) f is differentiable on (-5, 5)

∴ By Mean Value Theorem, there exists c ∈ (-5, 5) such that

⇒

⇒ 10 f'(c) = f(5) – f(-5)

It is given that f'(x) does not vanish anywhere.

∴ f'(c) ≠ 0

10 f'(c) ≠ 0

f(5) – f(-5) ≠0

f(5) ≠ f(-5)

Hence proved.

By Mean Value Theorem, it is proved that f(5) ≠ f(-5).

Given: f(x) = x² – 4x – 3 in the interval [1, 4]

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

As f(x) is a polynomial function,

(a) f(x) is continuous in [1, 4]

(b) f'(x) = 2x – 4

So, f(x) is differentiable in (1, 4).

∴

f(4) = 4² – 4(4) – 3 = 16 – 16 – 3 = -3

f(1) = 1² – 4(1) – 3 = 1 – 4 – 3 = -6

∴ There is a point c ∈ (1, 4) such that f'(c) = 1

⇒ f'(c) = 1

⇒ 2c – 4 = 1

⇒ 2c = 1+4 =5

⇒ c = 5/2 where c ∈ (1,4)

The Mean Value Theorem is verified for the given f(x).

Given: f(x) = x³ – 5x² – 3x in the interval [1, 3]

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

As f(x) is a polynomial function,

(a) f(x) is continuous in [1, 3]

(b) f'(x) = 3x² - 10x - 3

So, f(x) is differentiable in (1, 3).

∴

f(3) = 3³ – 5(3)² – 3(3) = 27 – 45 - 9 = -27

f(1) = 1³ – 5(1)² – 3(1)= 1 – 5 - 3 = -7

⇒

∴ There is a point c ∈ (1, 4) such that f'(c) = -10

⇒ f'(c) = -10

⇒ 3c² – 10c - 3 = -10

⇒ 3c² – 10c +7 =0

⇒ 3c² - 3c – 7c + 7 = 0

⇒ 3c(c – 1) – 7(c – 1) = 0

⇒(c – 1)(3c – 7) = 0

⇒ c = 1, 7/3 where c = 7/3 ∈ (1, 3)

The Mean Value Theorem is verified for the given f(x) and c = 7/3∈(1, 3) is the only point for which f'(c) = 0.

Mean Value Theorem states that for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

If a function does not satisfy any of the above conditions, then Mean Value Theorem is not applicable.

(i) f (x) = [x] for x ∈ [5, 9]

As the given function is a greatest integer function,

(a)f(x) is not continuous in [5, 9]

(b) Let y be an integer such that y ∈ (5, 9)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x=y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in [5, 9].

Here, f(x) does not satisfy the conditions of Mean Value Theorem.

Mean Value Theorem is not applicable for f(x) = [x] for x∈[5, 9].

(ii) f (x) = [x] for x ∈ [– 2, 2]

As the given function is a greatest integer function,

(a)f(x) is not continuous in [-2, 2]

(b) ) Let y be an integer such that y ∈ (-2, 2)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x=y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in (-2, 2)

Here, f(x) does not satisfy the conditions of Mean Value Theorem.

Mean Value Theorem is not applicable for f(x) = [x] for x∈[-2, 2].

(iii) f (x) = x² – 1 for x ∈ [1, 2]

As the given function is a polynomial function,

(a) f(x) is continuous in [1, 2]

(b) f'(x) = 2x

So, f(x) is differentiable in [1, 2].

Here, f(x) satisfies the conditions of Mean Value Theorem.

So, Mean Value Theorem is applicable for f(x).

∴

f(2) = 2² – 1 = 4 -1 = 3

f(1) = 1² – 1 = 1 – 1 = 0

⇒

∴ There is a point c ∈ (1, 2) such that f'(c) = 3

⇒ f'(c) = 3

⇒ 2c = 3

⇒ c = 3/2 where c ∈ (1, 2)

Mean Value Theorem is applicable for f(x) = x² – 1 for x∈[1, 2].

Let y = (3x² – 9x + 5)⁹

If u = v(w(x))

Then using chain rule

∴ Differentiating y w.r.t. x using chain rule

= 9(3x² – 9x + 5)⁸×(6x – 9)

=9(3x² – 9x + 5)⁸ × 3(2x – 3)

=27(3x² – 9x + 5)⁸ (2x – 3)

∴

Let y = sin³ x + cos⁶ x

Differentiating both sides with respect to x

∴ ∵ ]

= 3 sin² x × cos x + 6 cos⁵ x × ( – sin x)

= 3 sinx cosx (sinx– 2 cos⁴ x)

∴ = 3 sinx cosx (sinx– 2 cos⁴ x)

Let y = (5x)^{3cos 2x}

Then log y = log (5x)^{3cos 2x}

⇒ log y = 3cox 2x × log 5x

Differentiating both sides with respect to x, we get

[∵ ]

⇒

⇒

⇒

⇒

∴

Let y =

Differentiating both sides with respect to x, we get

Using chain rule we get

⇒

⇒

⇒

⇒

∴

Let y =

Differentiating both sides with respect to x, we get

Using Quotient rule

∴

Let y =

⇒

Substituting the value of in y.

∴ y =

⇒ y = x/2

Differentiating both sides with respect to x, we get

∴

Let y =

Taking logarithm on both sides

⇒ log y = log (log x)^{log x} = log x × log (log x)

Differentiating both sides with respect to x, we get

⇒

⇒

⇒

∴

Let y = cos (a cos x + b sin x)

a and b are some constants

y = cos (a cos x + b sin x)

Differentiating both sides with respect to x, we get

Using chain rule

⇒

⇒

⇒

∴

Let y =

Taking logarithm both sides, we get

log y = log [(sin x – cos x)^{(sin x – cos x)}]

⇒ log y = (sin x – cos x) × log (sin x – cos x)

Differentiating both sides with respect to x, we get

⇒

⇒

⇒

∴

Let y = x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0

And let x^x = u, x^a = v, a^x = w and a^a = s

Then y = u + v + w + s

∴ ……..(I)

Now,

u = x^x

Taking logarithm both sides, we get

log u = log x^x

⇒ log u = x log x

Differentiating both sides w.r.t. x

⇒

⇒ …….(II)

v = x^a

Differentiating both sides with respect to x

⇒ ………..(III)

w = a^x

Taking logarithm both sides

log w = log a^x

log w = x log a

Differentiating both sides with respect to x

⇒

⇒ …………….(IV)

s = a^a

Differentiating both sides with respect to x

… ……….(V)

Putting (II), (III), (IV) and (V) in (I)

∴

Let y =

And let = u & = v

∴ y = u + v

Differentiating both sides w.r.t. x we get

………..(I)

Now,

Taking logarithm both sides

⇒

Differentiating w.r.t. x, we get

⇒

⇒ ……………(II)

Also,

Taking logarithm both sides

⇒

Differentiating both sides w.r.t. x

⇒

⇒

⇒ ……………………….(II)

Substituting (II) and (III) in (I)

∴

To find we need to find out and

So,

Given, y = 12 (1 – cos t) and x = 10 (t – sin t)

x = 10 (t – sin t)

Differentiating with respect to t.

⇒

y = 12 (1 – cos t)

Differentiating with respect to t.

⇒

∴

∴

Given,

Differentiating with respect to x

⇒

⇒

⇒

⇒

⇒

⇒

∴

Given,

Now, squaring both sides, we get

⇒

⇒

⇒ x² + x²y = y² + y²x

⇒ x² – y² = xy² – x²y

⇒ (x + y)(x – y) = xy (y – x)

⇒ x + y = –xy

⇒ y + xy = –x

⇒ y (1 + x) = –x

⇒

Differentiating both sides with respect to x, we get

Using Quotient Rule

∴

Hence, Proved

Given, (x – a)² + (y – b)² = c²

Differentiating with respect to x, we get

⇒

⇒

∴

Differentiating again with respect to x

Using Quotient Rule

⇒

⇒

Substituting the value of dy/dx in the above equation

⇒

⇒

∴

∴ , which is independent of a and b

Hence, Proved

Given, cos y = x cos (a + y)

Differentiating both sides with respect to x

⇒

⇒

⇒ …………..(I)

Since, cos y = x cos (a + y) ⇒ x = cos y/cos (a + y)

Substituting the value of x in (I)

⇒

⇒

⇒

⇒

Hence, proved

Given, x = a (cos t + t sin t) and y = (sin t – t cos t)

To find we need to find out and

So, and

x = a (cos t + t sin t)

Differentiating with respect to t.

⇒

⇒

y = a (sin t – t cos t)

Differentiating with respect to t.

⇒

∴

Differentiating dy/dx with respect to t

And

⇒

∴

When, x ≥ 0,

f(x) = |x|³ = x³

So, f’(x) = 3x²

And f’’(x) = d(f’(x))/dx = 6x

∴ f’’(x) = 6x

When x < 0,

f(x) = |x|³ = ( – x)³ = – x³

f’(x) = – 3x²

f’’(x) = – 6x

∴

To prove : P(n) : = nx^{n – 1} for all positive integers n

For n = 1,

LHS = = 1

RHS = 1 × x^{1 – 1} = 1

So, LHS = RHS

∴ P(1) is true.

∴ P(n) is true for n = 1

Let P(k) be true for some positive integer k.