Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(a)

I. Compute the following limits.

Question 1.
Find the derivatives of the following func-tions f(x). 7286883416
i) \(\sqrt{x}+2 x^{\frac{3}{4}}+3 x^{\frac{5}{6}}\) (x > 0)
Solution:
y = \(\sqrt{x}+2 x^{\frac{3}{4}}+3 x^{\frac{5}{6}}\) (x>0)
\(\frac{dy}{dx}\) = \(\frac{1}{2}\).x-1/2 + 2.\(\frac{3}{4}\).x-1/4 + 3.\(\frac{5}{6}\).x-1/6 dx
= \(\frac{1}{2}\)[x-1/2 + 3.x-1/4 + 5.x-1/6]

ii) \(\sqrt{2 x-3}+\sqrt{7-3 x}\).
Solution:
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 1

iii) (x² – 3) (4x³ + 1)
Solution:
y = (x² – 3) (4x³ + 1)
\(\frac{dy}{dx}\) = (x² – 3) \(\frac{d}{dx}\) (4x³ + 1) + (4x³ + 1) \(\frac{d}{dx}\)(x² – 3)
= (x² – 3) (12x²) + (4x³ + 1) (2x)
= 12x4 – 36x² + 8x4 + 2x
= 20x4 – 36x² + 2x

iv) (√x – 3x) (x + \(\frac{1}{x}\))
Solution:
y = (√x – 3x) (x + \(\frac{1}{x}\))
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 2

v) (√x + 1) (x² – 4x + 2) (x > 0)
Solution:
y = (√x + 1) (x² – 4x + 2) (x > 0)
Differentiating w.r.to x
\(\frac{dy}{dx}\) = (√x + 1) \(\frac{d}{dx}\)(x² – 4x + 2) + (x² – 4x + 2) \(\frac{d}{dx}\)(√x +1)
= (√x + 1) (2x – 4) + \(\frac{x^{2}-4 x+2}{2 \sqrt{x}}\)

vi) (ax + b)n. (cx + d)m.
Solution:
y = (ax + b)n. (cx + d)m
\(\frac{dy}{dx}\) = (ax + b)n \(\frac{d}{dx}\)(cx +d)m + (cx + d)m \(\frac{d}{dx}\)(ax + b)n
= (ax + b)n [m(cx + d)m-1. c] + (cx + d)m [n(ax + b)n-1. a]
= (ax + b)n-1 (cx + d)m-1 [cm (ax + b) + an (cx + d)]
= (ax + b)n (cx + d)m [\(\frac{an}{ax+b} + \frac{cm}{cx+d}\) ]

vii) 5 sin x + ex log x
Solution:
y = 5 sin x + ex. log x
\(\frac{dy}{dx}\) = 5 cos x + ex. \(\frac{d}{dx}\) (log x) + log x \(\frac{d}{dx}\)(ex)
= 5 cos x + ex. \(\frac{1}{x}\) + (log x) (ex)

viii) 5x + log x + x³ ex
Solution:
y = 5x + log x + x³ ex
\(\frac{dy}{dx}\) = 5x . log 5 + \(\frac{1}{x}\) + x³.ex + ex.3x²
= 5x. log 5 + – + x³ ex + 3x² ex

ix) ex + sin x cos x
Solution:
y = ex + sin x . cos x
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) (ex) + \(\frac{d}{dx}\) (sin x . cos x)
= ex + sin x \(\frac{d}{dx}\) (cos x) + cos x \(\frac{d}{dx}\) (sin x)
= ex – sin² x + cos² x
= ex + cos 2x

x) \(\frac{p x^{2}+e x+r}{a x+b}\)(|a| + |b| ≠ 0)
Solution:
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 3
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 4

xi) log7 (log x) (x > 0)
Solution:
y = log7 (log x) (x > 0)
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 5

xii) \(\frac{1}{a x^{2}+b x+c}\) (|a| + |b| + |c| ≠ 0)
Solution:
\(\frac{1}{a x^{2}+b x+c}\) (|a| + |b| + |c| ≠ 0)
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 6

xiii) e2x log (3x + 4) (x > –\(\frac{4}{3}\))
Solution:
y = e2x. log (3x + 4) (x > –\(\frac{4}{3}\))
Differentiating w.r.to x
\(\frac{dy}{dx}\) = e2x \(\frac{d}{dx}\)[log (3x + 4) + log (3x + 4) \(\frac{d}{dx}\) (e2x)]
= e2x.\(\frac{1}{3x+4}\) 3 + log (3x + 4). e2x . 2
= e2x (\(\frac{3}{3x+4}\) + 2 log (3x + 4))

xiv) (4 + x²) e²xy
Solution:
y = (4 + x²). e2x
Differentiating w.r.to x
\(\frac{dy}{dx}\) = (4 + x²) \(\frac{d}{dx}\) (e2x) + e2x \(\frac{d}{dx}\)(4 + x²)
= (4 + x²). 2e2x + e2x (0 + 2x)
= 2e2x [4 + x² + x]
= 2e2x (x² + x + 4)

xv) \(\frac{ax+b}{cx+d}\) [|c| + |d|≠0]
Solution:
y = \(\frac{ax+b}{cx+d}\) [|c| + |d|≠0]
Differentiating w.r.to x
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 7

xvi) ax. e
Solution:
y = ax. e
Differentiating w.r.to x
\(\frac{dy}{dx}\) = (ax) \(\frac{d}{dx}\)(e) + (e)\(\frac{d}{dx}\)(ax)
= ax. e. 2x + e. ax. log a
= ax. (2x + log a)
= y(2x + log a)

Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a)

Question 2.
If f(x) = 1 + x + x² + + x100, then find f’ (1).
Solution:
f'(x) = 1 + 2x + 3x² + 100 x99
f'(1) = 1+2 + 3 ….+ 100
= \(\frac{100 \times 101}{2}=5050\left(\Sigma x=\frac{x(x+1)}{2}\right)\)

Question 3.
If f (x) = 2x² + 3x – 5, then prove that f(0) + 3f (-1) = 0.
Solution:
f'(x) = 4x + 3
f'(0) = 0 + 3 = 3
f'(-1) = – 4 + 3 = -1
f'(0) + 3f'(-1) = 3 + 3(-1) = 3 – 3 = 0 n.

II.

Question 1.
Find the derivatives of the following functions from the first principles.
i) x³
Solution:
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 8
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 9
= 3x² + 0 + 0 = 3x²

ii) x4 + 4
Solution:
f(x + h) – f(x) = ((x + h)4 + 4) – (x4 + 4)
= (x + h)4 + 4 – x4 – 4 .
= x4 + 4x³h + 6x²h² + 4xh³ + h4 – x4
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 10
= 4x³ + 0 + 0 + 0 = 4x³

iii) ax² + bx + c
Solution:
f(x + h) = a(x + h)² + b(x + h) + c
= a(x² + 2hx + h²) + b(x + h) + c
= ax² + 2ahx + ah² + bx + bh +c

f(x + h) – f(x) = ax² + 2ahx + ah² + bx + bh + c – ax² – bx – c
= h [2ax + ah + b]
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 11
= 2ax + 0 + b = 2ax + b

iv) \(\sqrt{x+1}\)
Solution:
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 12

v) sin 2x
Solution:
f(x + h) – f(x) = sin 2(x + h) – sin 2x
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 13

vi) cos ax
Solution:
f(x + h) – f(x) = cos a (x + h) – cos ax
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 14
= – 2 sin ax. \(\frac{a}{2}\)
=-a. sin ax

vii) tan 2x
Solution:
f(x + h) – f(x) = tan 2(x + h) – tan 2x
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 15

viii) cot x
Solution:
f(x + h) – f(x) = cot (x + h) – cot x
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 16
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 17

ix) sec 3x
Solution:
f(x + h) – f(x)= sec 3(x + h) – sec 3x
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 18

x) x sin x
Solution:
f(x + h) – f(x) = (x + h) sin (x + h) – x sin x
= x (sin (x + h) – sin x) + h. sin (x + h)
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 19

xi) cos² x
Solution:
f(x + h) – f(x) = cos² (x + h) – cos² x
= -(cos² x – cos² (x + h))
= -sin (x + h + x) sin (x + h – x)
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 20
= -sin 2x. 1 = -sin 2x

Question 2.
Find the derivatives of the following function.
i) \(\frac{1-x \sqrt{x}}{1+x \sqrt{x}}\) (x > 0)
Solution:
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 21

ii) xn nx log (nx) (x > 0, n ∈ N)
Solution:
y = xn. nx. log nx
\(\frac{dy}{dx}\) = xn. nx (log nx) + nn. log xn (xn) + xn. log nx (nx)
= xn. nx \(\frac{n}{log nx}\) + nx. log xn (nxn-1) + xn . log nx. (nx . log nx)
= xn-1. nx[\(\frac{nx}{log nx}\) + log nx. (nn. log nx)]

iii) ax2n. log x + bxn e-x
Solution:
y = ax2n. log x + bxn e-x
\(\frac{dy}{dx}\) = a (x2n.\(\frac{1}{x}\) + log x (2nx2n-1)) + b n (-e-x) + e-x. nxn-1)
= a. x2n-1 + 2an. x2n-1. log x – bxn e-x + bn. xn-1 . e-x

iv) (\(\frac{1}{x}\) – x)³ ex
Solution:
y = (\(\frac{1}{x}\) – x)³ . ex
\(\frac{dy}{dx}\) = (\(\frac{1}{x}\) – x)³ \(\frac{d}{dx}\)(ex) + ex \(\frac{d}{dx}\){(\(\frac{1}{x}\) – x)³}
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 22

Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a)

Question 3.
Show that the function f(x) = |x| + |x – 1|, x ∈ R is differentiable for all real numbers except for 0 and 1.
Solution:
f(x) = |x| + |x – 1| ∀ x ∈ R
f(x) = x + x – 1 = 2x – 1, x ≥ 1
= x – (x – 1) = x – x + 1, = 1, 0 < x < 1
= -x – (x – 1) =-x – x + 1 = 1 – 2x, x < 0 ∴ f(x) = 2x – 1, x > 1
= 1, 0 < x < 1 = 1 – 2x, x ≤ 0 If x > 1, then f(x) = 2x – 1 = polynomial in x f(x) is differentiable for all x > 1
If 0 < x < 1, then f(x) = 1 – constant
∴ f(x) is differentiable if 0 < x < 1.
If x < 1, then f(x) = 1 – 2x = polynomial in x.
∴ f(x) is differentiable for all x < 1

Case (i) : x = 0
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 23
R f'(0) ≠ Lf'(0)
∴ f'(0) does not exist.
f(x) is not differentiable at x = 0.

Case (ii): x = 1
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 24
R f'(1) ≠ L f'(1)
f(1) does not exist.
f(x) is not differentiable at x = 1
∴ f(x) is differentiable for all real x except zero and one.

Question 4.
Verify whether the following function is differentiable at 1 and 3.
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 25
Solution:
Case (i): x = 1
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 26
R f'(1) ≠ L f'(1)
f(x) is not differentiable at x = 1

Case (ii) : x = 3
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 27
f(3+) ≠ f'(3)
f(x) is not differentiable at x = 3.

Question 5.
Is the following function f derivable at 2?
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 28
Solution:
Inter 1st Year Maths 1B Differentiation Solutions Ex 9(a) 29
f'(2) ≠ f(2+); f(x) is not derivable at x = 2.

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