# Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Miscellaneous Exercise on Chapter 13 | Set 1

### Question 1: Find the derivative of the following functions from first principle:

### (i) -x

**Solution:**

f(x) = -x

f(x+h) = -(x+h)

From the first principle,

f'(x) = -1

### (ii) (-x)^{-1}

**Solution:**

f(x) = (-x)

^{-1}=f(x+h) = (-(x+h))

^{-1 }=From the first principle,

### (iii) sin(x+1)

**Solution:**

f(x) = sin(x+1)

f(x+h) = sin((x+h)+1)

From the first principle,

Using the trigonometric identity,

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

f'(x) = cos (x+1) (1)

f'(x) = cos (x+1)

### (iv)

**Solution:**

Here,

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sinsinMultiplying and dividing by 2,

### 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 2: (x+a)

**Solution:**

f(x) = x+a

Taking derivative both sides,

As, the derivative of x

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

### Question 3:

**Solution:**

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’+u’vAs, the derivative of x

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

### Question 4: (ax+b) (cx+d)^{2}

**Solution:**

f(x) = (ax+b) (cx+d)

^{2}Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’+u’vAs, the derivative of x

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

### Question 5:

**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 6:

**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 7:

**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 8:

**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 9:

**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 10:

**Solution:**

Taking derivative both sides,

As, the derivative of x

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

### Question 11:

**Solution:**

Taking derivative both sides,

As, the derivative of x

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

### Question 12: (ax+b)^{n}

**Solution:**

f(x) = (ax+b)

^{n}f(x+h) = (a(x+h)+b)

^{n}f(x+h) = (ax+ah+b)

^{n}From the first principle,

Using the binomial expansion, we have

### Question 13: (ax+b)^{n} (cx+d)^{m}

**Solution:**

f(x) = (ax+b)

^{n}(cx+d)^{m}Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’+u’v

Let’s take, g(x) = (cx+d)

^{m}g(x+h) = (c(x+h)+d)

^{m}g(x+h) = (cx+ch+d)

^{m}From the first principle,

Using the binomial expansion, we have

So, as

### Question 14: sin (x + a)

**Solution:**

f(x) = sin(x+a)

f(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

### Question 15: cosec x cot x

**Solution:**

f(x) = cosec x cot x

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’+u’vf'(x) = cot x (-cot x cosec x) + (cosec x) (-cosec

^{2}x)f'(x) = – cot

^{2}x cosec x – cosec^{3}x