Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 1.

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

Solution:

Derivative of f(x) at x = a is given by

f(a) = \(\lim _{h \rightarrow 0}\) \(\frac{f(a+h)-f(a)}{h}\)

Here, f(x) = x^{2} – 2

a = 10, âˆ´ f(10 + h) = (10 + h)^{2} – 2

and f(10) = 10^{2} – 2.

âˆ´ f(10 + h) – f(10) = [(10 + h)^{2} – 2] – (10^{2} – 2)

= (10 + h)^{2} – 10^{2} = (20 + h)h

âˆ´ f(10) = \(\lim _{h \rightarrow 0}\) \(\frac{f(10+h)-f(10)}{h}\)

= \(\lim _{h \rightarrow 0}\) \(\frac{h(20+h)}{h}\).

= \(\lim _{h \rightarrow 0}\) (20 + h) = 20.

Question 2.

Find the derivative of 99x at x =100.

Solution:

Derivative of f(x) at x = 100 is

\(\lim _{h \rightarrow 0}\) = \(\frac{f(100+h)-f(100)}{h}\)

Now, f(x) = 99x.

âˆ´ f(100 + h) = 99(100 + h).

and f(100) = 99 Ã— 100.

âˆ´ f(100 + h) – f(100) = 99(100 + h) – 99 Ã— 100

= 99[100 + h – 100] = 99 Ã— h

âˆ´ f(100) = \(\lim _{h \rightarrow 0}\) \(\frac{99h}{h}\) = 99.

Question 3.

Find the derivative of x at x = 1.

Solution:

Derivative of f(x) = x at x = 1

f'(1) = \(\lim _{h \rightarrow 0}\) = \(\frac{f(x+h)-f(x)}{h}\).

= \(\lim _{h \rightarrow 0}\) \(\frac{(1+h)-1}{h}\)

= \(\lim _{h \rightarrow 0}\) \(\frac{h}{h}\) = 1.

Question 4.

Find the derivative of the following functions from first principles:

(i) x^{3} – 27

(ii) (x – 1)(x – 2)

(iii) \(\frac{1}{x^{2}}\)

(iv) \(\frac{x+1}{x-1}\)

Solution:

(i) f(x) = x^{3} – 27, Derivative f'(x) by first principle is given by

= 3x^{2}

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

(iii) f(x) = \(\frac{1}{x^{2}}\).

Derivative of f(x) is given by

(iv) f(x) = \(\frac{x+1}{x-1}\).

Question 5.

For the function f(x) = \(\frac{x^{100}}{100}\) + \(\frac{x^{99}}{99}\) + …………… + \(\frac{x^{2}}{2}\) + x, prove that f ‘(1) = 100f ‘(0).

Solution:

Now, f ‘(1) = 1 + 1 + ………… to 100 terms = 100

Also, f ‘(0) = 1.

âˆ´ f ‘(1) = 100 Ã— 1 = 100 f ‘(0)

Hence, f ‘(1) = 100 f ‘(0).

Question 6.

Find the derivative of x^{n} + ax^{n-1} + a^{2}x^{n-2} + …………. + a^{n-1}x + a^{n} for some fixed real number.

Solution:

âˆ´ f(x) = nx^{n-1} + (n – 1)ax^{n-2} + (n – 2)a^{2}x^{n-3} + …………. + a^{n-1}.

Question 7.

For some constants a and b, find the derivative of:

(i) (x – a)(x – b)

(ii) (ax^{2} + b)^{2}

(iii) \(\frac{x-a}{x-b}\)

Solution:

(i) f(x) = (x – a)(x – b)

We have:

f'(uv) = u’v + uv’.

âˆ´ f'(x) = x – b + x – a = 2x – a – b.

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

Now, \(\frac{d}{dx}\) x^{4} = 4x^{3} and \(\frac{d}{dx}\) x^{2} = 2x and \(\frac{d}{dx}\) b^{2} = 0.

âˆ´ f(x) = a^{2}.4x^{3} + 2ab.2x + 0

= 4a^{2}x^{3} + 4abx.

(iii)

Question 8.

Find the derivative of \(\frac{x^{n}-a^{n}}{x-a}\) for some constant a.

Solution:

Question 9.

Find the derivative of

(i) 2x – \(\frac{3}{4}\)

(ii) (5x^{3} + 3x – 1)(x – 1)

(iii) x^{-3}(5 + 3x)

(iv) x^{5}(3 – 6x^{-9})

(v) x^{-4}(3 – 4x^{-5})

(vi) \(\frac{2}{x+1}\) – \(\frac{x^{2}}{3 x-1}\)

Solution:

(i) Let f(x) = 2x – \(\frac{3}{4}\).

âˆ´ f ‘(x) = 2.1 [âˆµ \(\frac{d}{dx}\) x = 1, \(\frac{d}{dx}\) (\(\frac{3}{4}\)) = 0]

= 2

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

(uv)’ = u’v + uv’

âˆ´ f(5x^{3} + 3x – 1)(x – 1)

= [\(\frac{d}{dx}\) (5x^{3} + 3x – 1)] (x – 1) + (5x^{3} + 3x – 1) \(\frac{d}{dx}\) (x – 1)

= (15x^{2} + 3)(x – 1) + (5x^{3} + 3x – 1).1

= x(15x^{2} + 3) – (15x^{2} + 3) + (5x^{3} + 3x – 1)

= 20x^{3} – 15x^{2} + 6x – 4.

(iii) Let f(x) = x^{-3}(5 + 3x) = 5x^{-3} + 3x^{-2}

âˆ´ f ‘(x) = 5(- 3) x^{-4} + 3(- 2)x^{-3}

(iv) Let f(x) = x^{5}(3 – 6x^{-9}) = 3x^{5} – 6x^{-4}

f ‘(x) = 3 Ã— 5x^{4} – 6 Ã— (- 4)x^{-5}

= 15x^{4} + 24x^{-5}

(v) Let f(x) = x^{-4}(3 – 4x^{-5}) = 3x^{-4} – 4x^{-9}

f ‘(x) = 3.(- 4)x^{-5} – 4. (- 9)x^{-10}

(vi) Let f(x) = \(\frac{2}{x+1}\) – \(\frac{x^{2}}{3 x-1}\)

Question 10.

Find the derivative of cos x from the first principle.

Solution:

f(x) = cos x.

âˆ´ By first principle,

Question 11.

Find the derivative of the following functions:

(i) sin x cos x

(ii) sec x

(iii) 5secx + 4cos x

(iv) cosec x

(v) 3cot x + 5cosec x

(vi) 5sinx – 6cosx + 7

(vii) 2tanx – 7 secx

Solution:

(i) Let f(x) = sin x.cos x

So, f ‘(x) = u’v + uv’

= (\(\frac{d}{dx}\) sin x)cos x + sin x \(\frac{d}{dx}\)(cos x)

= cos x. cos x + sin x(- sin x)

= cos^{2}x – sin^{2}x = cos 2x.

(ii)

= sec x. tan x

Alternative method:

(vi) \(\frac{d}{dx}\) sin x = cos x and \(\frac{d}{dx}\) cos x = – sin x.

âˆ´ \(\frac{d}{dx}\) (5sin x – 6cos x + 7) = 5cos x + 6sin x = 0

= 5cos x + 6sin x.

(vii) \(\frac{d}{dx}\) (sec x) = sec x tan x

and \(\frac{d}{dx}\)(tan x) = sec^{2}x

âˆ´ \(\frac{d}{dx}\)(2tan x – 7 secx) = 2sec^{2}x – 2 sec x tan x

= sec x(2 sec x – 7 tan x).