## Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.3

Question 1.

Find the derivative of x^{2} – 2 at x = 10.

Solution:

Let f(x) = x^{2} – 2

Question 2.

Find the derivative of 49x at x = 50.

Solution:

Let f(x) = 49x

Question 3.

Find the derivative of the following function from first principle:

Solution:

(i) Let y = x^{3} – 16

Again, let y + δy = (x + δx)^{3} – 16

⇒ δy = (x + δx)^{3} – 16 – y

⇒ δy = (x + δx)^{3} – 16 – x^{3} + 16

⇒ δy = (x + δx)^{3} – x^{3}

(ii) Let y = (x – 1) (x – 2) = x^{2} – 3x + 2

Again, let y + δy = (x + δx)2 – 3(x + δx) + 2

⇒ δy = (x + δx)^{2} – 3(x + δx) + 2 – x^{2} + 3x – 2

Question 4.

For the function

Prove that f'(1) = 100 f'(0).

Solution:

Then, putting 1 and 0 in place of x.

f'(1)= (1^{99} + 1^{98} + … + 1)+ 1

= 1 + 1 + 1 + …+ 99 term + 1

= 99+ 1 = 100 and f'(0) = 1

Hence, f'(1)= 100

∵ f'(1) = 100 f'(0) Hence Proved.

Question 5.

For any constant real number a, find the derivative of:

x^{n} + ax^{n – 1} + a^{2}x^{n – 2} + … + a^{n – 1} x + a^{n}

Solution:

Let y =f(x) = x^{n} + ax^{n – 1} + a^{2}x^{n – 2} + …… + a^{n – 1}x + a^{n}

Then, derivative of f(x),

Question 6.

For some constant a and b, find the derivative of the following functions :

Solution:

(i) Let y = f(x) = (x – a) (x – b) or y = f(x) = x^{2} – (a + b)x + ab

Then, derivative of given function

Hence, derivative of given function (x – a) (x – b)

= 2x – a – b

(ii) Let y = f(x) = (ax^{2} + b)^{2}

or y = f(x) = a^{2}x^{4}+ 2abx^{2} + b^{2}

Then, derivative of given function

= 4a^{2}x^{3} + 4abx = 4ax(ax^{2} + b)

Hence, derivative of given function (ax^{2} + b^{2})^{2}

= 4a^{2}x^{3} + 4abx or 4ax(ax^{2} + b)

We know that if any function is in the form of fraction, then its derivative

Question 7.

For any constant a, find the derivative of

.

Solution:

Question 8.

Find the derivative of the following 3

Solution:

(ii) Let y = f(x) = (5x^{3} + 3x – 1) (x – 1)

The given function is product of two function.

Then, derivative of product of two functions

= 20x^{3} – 15x^{2} + 6x – 4

Hence, derivative of given function = 20x^{3} – 15x^{2} + 6x – 4.

(iii) Let y = x^{5}(3 – 6x^{-9})

Then, derivative of given function

Hence, derivative of given function

We can also solve this equation by product rule of derivative.

Question 9.

Find the derivative of cos x by first principle.

Solution:

Let

f(x) = cos x, then f(x + h) = cos(x + h)

Then

Question 10.

Find the derivatives of the following :

(i) sin x cos x

(ii) sec x

(iii) cosec x

(iv) 3 cot x + 5 cosec x

(v) 5 sin x – 6 cos x + 7

Solution:

(i) Let f(x) = sin x. cos x, which is product of two functions.

So, formula of derivative of product of two functions.

= – sin^{2} x + cos^{2} x

= cos^{2} x – sin^{2} x

= cos 2x ( ∵ cos^{2} x – sin^{2} x = cos^{2}x)

Hence, derivative of given function sin x cos x = cos 2x

(ii) Let f(x) = sec x

Hence, derivative of the given function sec x = sec x tan x

(iii) Let f(x) = cosec x

Then, derivative of f(x)

= – cosec x cot x

Hence, derivative of the given function cosec x

= – cosec x cot x

(iv) Let f(x) = 3 cot x + 5 cosec x

Hence, derivative of the given function 3 cot x + 5 cosec x is – 3 cosec^{2} x – 5 cosec x cot x

(v) Let f(x) = 5 sin x – 6 cos x + 7

Hence, derivative of the given function 5 sin x – 6 cos x + 7 is 5 cos x + 6 sin x.