### Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

### Question 16:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

### Question 17:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

### Question 18:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

### Question 19: sin^{n} x

**Solution:**

f(x) = sin

^{n}x

When n = 1,f(x) = sin x

When n = 2,f(x) = sin

^{2}x = sin x sin xUsing the product rule, we have

(uv)’ = uv’+vu’f'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

f(x) = sin

^{3}x = sin^{2}x sin xUsing the product rule, we have

(uv)’ = uv’+vu’Pattern w.r.t n is seen here, as follows

Let’s check this statement.

For P(n) = n sin

^{n-1}x cos xFor P(1),

P(1) = 1 sin

^{1-1}x cos x = cos x. Which is true.n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’= (sin

^{k}x) + (sin x)= (sin

^{k}x) (cos x) + (sin x) (k sin^{k-1 }x cos x)= (sin

^{k}x) (cos x)[k+1]Hence proved for P(k+1).

So, is true.

### Question 20:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.

### Question 21:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

Let’s take g(x) = sin (x+a)

g(x+h) = sin((x+h)+a)

From the first principle,

Using the trigonometric identity,

sin A – sin B = 2 cossinMultiply and divide by 2, we have

Hence,

Using the trigonometric identity,

cos A cos B + sin A sin B = cos (A-B)

### Question 22: x^{4}(5sin x – 3cos x)

**Solution:**

f(x) = x

^{4}(5sin x – 3cos x)Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.f'(x) = (x

^{4}) [(5 cos x) – (3 (- sin x))] + (5sin x – 3cos x)(4x^{3})f'(x) = (x

^{4}) [(5 cos x) + (3 sin x)] + (5sin x – 3cos x)(4x^{3})f'(x) = (x

^{3}) [5x cos x + 3x sin x + 20sin x – 12 cos x]

### Question 23: (x^{2}+1) cos x

**Solution:**

f(x) = (x

^{2}+1) cos xTaking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.f'(x) = (x

^{2}+1) (- sin x) + (cos x)[(2x^{2-1})+0]f'(x) = -x

^{2}sin x- sin x + 2x cos x

### Question 24:

**Solution:**

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.

### Question 25: (x + cos x)(x – tan x)

**Solution:**

f(x) = (x + cos x)(x – tan x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)g'(x) = sec

^{2}xHence,

f'(x) = (x + cos x) + (x – tan x)[1 + (- sin x)]

f'(x) = (x + cos x) [1 – (sec

^{2}x)] + (x – tan x)[1 – sin x]f'(x) = (x + cos x) [tan

^{2}x] + (x – tan x)[1 – sin x]f'(x) = tan

^{2}x(x + cos x) + (x – tan x)[1 – sin x]

### Question 26:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.

### Question 27:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.

### Question 28:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)g'(x) = sec

^{2}xHence,

### Question 29: (x + sec x) (x-tan x)

**Solution:**

f(x) = (x + sec x) (x-tan x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’As, the derivative of x

^{n}is nx^{n-1 }and derivative of constant is 0.Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)g'(x) = sec

^{2}xNow, let’s take h(x) = sec x =

h(x+h) =

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sinsinMultiply and divide by 2, we have

Hence,

### Question 30:

**Solution:**

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of x

^{n}is nx^{n-1}and derivative of constant is 0.Let’s take, g(x) = sinn x

When n = 1,

g(x) = sin x

When n = 2,

Using the product rule, we have

(uv)’ = uv’+vu’g'(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

g(x) = sin

^{3}x = sin^{2}x sin xUsing the product rule, we have

(uv)’ = uv’+vu’Pattern w.r.t n is seen here, as follows

Let’s check this statement.

For

For P(1),

. Which is true.

n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’Hence proved for P(k+1).

So, is true.

So, the given equation will be