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 cos  sin 

Multiply 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 sin  sin 

Multiplying 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 xⁿ 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’v

As, the derivative of xⁿ 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’v

As, the derivative of xⁿ 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 xⁿ 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 xⁿ 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 xⁿ 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 xⁿ 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 xⁿ is nx^n-1 and derivative of constant is 0.

Question 10: 

Solution:

Taking derivative both sides,

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Question 11: 

Solution:

Taking derivative both sides,

As, the derivative of xⁿ 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 cos  sin 

Multiply 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’v

f'(x) = cot x (-cot x cosec x) + (cosec x) (-cosec² x)

f'(x) = – cot² x cosec x – cosec³ x