TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(c)

Students must practice these TS Inter 1st Year Maths 1B Study Material Chapter 9 Differentiation Ex 9(c) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Differentiation 9(c)

I.
Question 1.
Find the derivatives of the following functions.
(i) sin^-1(3x – 4x³) (May 2011) (V.S.A.Q.)
Answer:
Let y = sin^-1(3x – 4x³)
Let x = sin θ then y = sin^-1(3 sin θ – 4 sin³θ)
= sin^-1 (sin 3θ) = 3θ = 3 sin^-1x
∴ \(\frac{d y}{d x}\) = 3 \(\frac{d}{d x}\) (sin^-1x) = \(\frac{3}{\sqrt{1-x^2}}\)

(ii) cos^-1 (4x³ – 3x) (March 2014) (V.S.A.Q.)
Answer:
Let y = cos^-1 (4x³ – 3x)
Suppose x = cos θ, then y
= cos^-1 (4 cos³ θ – 3 cos θ)
= cos^-1 (cos 3θ) = 3θ = 3 cos^-1x
∴ \(\frac{d y}{d x}\) = 3 . \(\frac{d}{d x}\) (cos^-1x) = \(\frac{-3}{\sqrt{1-x^2}}\)

(iii) sin^-1\(\left(\frac{2 x}{1+x^2}\right)\) (S.A.Q.)
Answer:
Let y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Suppose x = cos θ
then y = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= sin^-1 (sin 2θ) = 2θ = 2 tan^-1x
∴ \(\frac{d y}{d x}\) = 2 . \(\frac{d}{d x}\) (tan^-1x) = \(\frac{2}{\sqrt{1+x^2}}\)

(iv) tan^-1 \(\left(\frac{\mathbf{a}-\mathbf{x}}{1+\mathbf{a x}}\right)\) (S.A.Q.)
Answer:

(v) tan^-1\(\sqrt{\frac{1-\cos x}{1+\cos x}}\) (S.A.Q.)
Answer:

(vi) sin [cos (x²)] (S.A.Q.)
Answer:
Let y = sin [cos (x²)]
Let x² = u, cos u = v and y = sin v

(vii) sec^-1 \(\left(\frac{1}{2 x^2-1}\right)\) (0 < x < \(\frac{1}{\sqrt{2}}\))
Answer:

(viii) sin [tan^-1 (e^-x)] (V.S.A.Q.)
Answer:
Let y = sin [tan^-1 (e^-x)]
Let e^-x = u, tan^-1u = v and y = sin v

Question 2.
Differentiate f(x) with respect to g(x) for the following. (S.A.Q.)
(i) f(x) = e^x, g(x) = √x
Answer:
Let y = e^x and z = √x

(ii) f(x) = e^{sin x}, g(x) = sin x (S.A.Q.)
Answer:
Let y = e^{sin x} and z = sin x

(iii) f(x) = tan^-1\(\left(\frac{2 x}{1-x^2}\right)\), g(x) = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Answer:

Question 3.
If y = e^{a sin-1 x} then prove that (V.S.A.Q.)
\(\frac{d y}{d x}=\frac{a y}{\sqrt{1-x^2}}\)
Answer:
y = e^{a sin-1 x}

II.
Question 1.
Find the derivatives of the following functions. (S.A.Q.)
(i) tan^-1\(\left(\frac{3 a^2 x-x^3}{a\left(a^2-3 x^2\right)}\right)\)
Answer:

(ii) tan^-1 (sec x + tan x)
Answer:
Let y = tan^-1 (sec x + tan x)

(iii) tan^-1\(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\)
Answer:

(iv) (log x)^{tan x}
Answer:
Let y = (log x)^{tan x}
Then log y = tan x lobg I(log x)
Differentiating both sides with respect to ‘x’,

(v) (x^x)²
Answer:
Let y = x^x2
∴ log y = x² log x

(vi) 20^{log (tan x)}
Answer:
Let y = 20^{log (tan x)}
Then log y = log(tan x) log 20
Differentiating with respect to ‘x’

(vii) x^x + e^ex
Answer:
Let u = x^x and v = e^ex

(viii) (x log x) log (log x)
Answer:
Let y = x . log x . log (log x)
Then
\(\frac{dy}{d x}\) = x.log x . \(\frac{d}{d x}\) [log (log x)] + log x . log(log x) \(\frac{d}{d x}\) (x) + x . log(log x). \(\frac{d}{d x}\) (log x)
= x log x . \(\frac{1}{x log x}\) + log x log(log x) . 1 + x . log(log x) \(\left(\frac{1}{x}\right)\)
= 1 + log x . log (log x) + log (log x)
= log_ee + log (log x) + log x . log (log x)
= log (e log x) + log x . log (log x)

(ix) e^-ax2 . sin (x log x)
Answer:
Let y = e^-ax2 . sin (x log x)
Then
\(\frac{d y}{d x}\) = e^-ax2 \(\frac{d}{d x}\) [sin(x log x)] + sin(x log x) \(\frac{d}{d x}\) [e^-ax2]
= e^-ax2 cos (x log x) [x \(\left(\frac{1}{x}\right)\) + log x] + sin (x log x) (e^-ax2) . (- 2ax)
= e^-ax2 [cos (x log x) (1 + log x) – sin (x log x) . 2ax]
= e^-ax2 [cos (x log x) (log xe) – 2ax sin (x log x)]

(x) sin^-1\(\left(\frac{2^{x+1}}{1+4^x}\right)\) (Put 2^x = tan θ)
Answer:

Question 2.
Find \(\frac{d y}{d x}\) for the following functions.
(i) x = 3 cos t – 2 cos³t
y = 3 sin t – 2 sin³t (E.Q.)
Answer:
\(\frac{d x}{d t}\) = – 3 sin t + 6 cos²t(sin t)
= 3 sin t(2 cos²t – 1) = 3 sin t cos 2t
\(\frac{d y}{d t}\) = 3 cos t – 6 sin²t cos t
= 3 cos t (1 – 2 sin²t) = 3 cos t . cos 2t

(ii) x = \(\frac{3 a t}{1+t^3}\), y = \(\frac{3 a t^2}{1+t^3}\) (E.Q.) [Board Model paper]
Answer:

(iii) x = a (cos t + t sin t),
y = a (sin t – t cos t). (S.A.Q.)
Answer:
\(\frac{d x}{d t}\) = a(- sin t + t cos t + sin t)
= at cos t
\(\frac{d y}{d t}\) = a [cos t – (- t sin t + cos t)] = at sin t

(iv) x = a\(\left[\frac{1-t^2}{1+t^2}\right]\), y = \(\frac{2 b t}{1+t^2}\) (S.A.Q.)
Answer:

Question 3.
DifferentIate f(x) with respect to g(x) for the following. (S.A.Q.)
(i) f(x) = log_a^x, g(x) = a^x
Answer:
Let y = log_a^x and z = a^x

(ii) f(x) = sec^-1\(\left(\frac{1}{2 x^2-1}\right)\), g(x) = \(\sqrt{1-x^2}\)
Answer:

(iii) f(x) = tan^-1\(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\), g(x) = tan^-1x [June 2009 IPE]
Answer:

Question 4.
Find the derivative of the function y defined implicitly, by each of the following equations. (S.A.Q.)
(i) x⁴ + y⁴ – a² xy = 0
Answer:
Differentiating with respect to ‘x

(ii) y = x^y (S.A.Q.) (March 2004)
Answer:
Given y = x^y
Then log y = y log x ………………. (1)
Differentiating w.r.t. to ‘x’ both sides

(iii) y^x = x^{sin y} (S.A.Q.)
Answer:
Taking logarithm on both sides
x log y = sin y . log x
Differentiating w.r.t to ‘x’

Question 5.
Establish the following.
(i) If \(\sqrt{1-x^2}\) + \(\sqrt{1-y^2}\) = a(x – y) then \(\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\) (E.Q.) (May 2014, March 2009)
Answer:
Take x = sin θ and y = sin Φ
∴ From the given condition

(ii) If y = x\(\sqrt{a^2+x^2}\) + a² log(x + \(\sqrt{a^2+x^2}\)) then \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = 2\(\sqrt{a^2+x^2}\) (E.Q.)
Answer:

(iii) If x^{log y} = log x then \(\frac{d y}{d x}\) = \(\left[\frac{1-\log x \log y}{(\log x)^2}\right]\) (S.A.Q)
Answer:
Given x^{log y} = log x
Taking logarithm on both sides
log y . log x = log (log x)
Differentiating with respect to ‘x’

(iv) If y = tan^-1\(\left(\frac{2 x}{1-x^2}\right)\) + tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) – tan^-1\(\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)\) then \(\frac{d y}{d x}=\frac{1}{1+x^2}\) (S.A.Q.)
Answer:
Let x = tan θ, then

= tan^-1(tan 2θ) + tan^-1 (tan 3θ) – tan^-1 (tan 4θ)
= 2θ + 3θ – 4θ = θ = tan^-1x
∴ \(\frac{d y}{d x}=\frac{1}{1+x^2}\)

(v) If x^y = y^x then \(\) (S.A.Q.)
Answer:
Given x^y = y^x
Then log x^y = log y^x
⇒ y log x = x log y
Differentiating both sides w.r.t to ‘x’

(vi) If x^{\(\frac{2}{3}\)} + y^{\(\frac{2}{3}\)} = a^{\(\frac{2}{3}\)} then \(\frac{d y}{d x}=-\sqrt[3]{\frac{y}{x}}\) (S.A.Q.)
Answer:
Given x^{\(\frac{2}{3}\)} + y^{\(\frac{2}{3}\)} = a^{\(\frac{2}{3}\)}
Then differentiating both sides w.r.t to ‘x’

Question 6.
Find \(\frac{d y}{d x}\) of each of the following functions. (S.A.Q.)

(i) y = \(\frac{(1-2 x)^{\frac{2}{3}}(1+3 x)^{\frac{-3}{4}}}{(1-6 x)^{\frac{5}{6}}(1+7 x)^{\frac{-6}{7}}}\)
Answer:

(ii) y = \(\frac{x^4 \sqrt[3]{x^2+4}}{\sqrt{4 x^2-7}}\)
Answer:

(iii) y = \(\frac{(a-x)^2(b-x)^3}{(c-2 x)^3}\)
Answer:
log y = log (a – x)² + log(b – x)³ – log (c – 2x)³
= 2 log (a – x) + 3 log (b – x) – 3 log (c – 2x)
Differentiating both sides w.r.t to ‘x’

(iv) y = \(\frac{x^3 \sqrt{2+3 x}}{(2+x)(1-x)}\)
Answer:
log y = log \(\left[\frac{x^3 \sqrt{2+3 x}}{(2+x)(1-x)}\right]\)
= log x³ + \(\frac{1}{2}\) log (2 + 3x) – log (2 + x) – log (1 – x)
= 3 log x + \(\frac{1}{2}\) log (2 + 3x) – log (2 + x) – log (1 – x)
Differentiating both sides w.r.t to ‘x’

(v) y = \(\sqrt{\frac{(x-3)\left(x^2+4\right)}{3 x^2+4 x+5}}\)
Answer:

III.
Question 1.
Find the derivatives of the following functions. (E.Q.) (March 2013)
(i) (sin x)^{log x} + x^{sin x}
Answer:
y = (sin x)^{log x} + x^{sin x}
Let y = u + v where u = (sin x)^{log x} ………….. (1)
and v = x^{sin x} ……………….. (2)
From (1) log u = log x log (sin x)
Differentiate w.r.to x both sides

(ii) x^xx
Answer:
Let y = x^xx
∴ log y = log x^(xx) = x^x log x
Differentiate both sides w.r.t. ‘x’.
\(\frac{1}{y} \frac{d y}{d x}\) = x^x \(\left(\frac{1}{x}\right)\) + log x . x^x (1 + log x)
(∵ \(\frac{d y}{d x}\) (x^x) = x^x (1 + log x))
= x^{x – 1} + x^x . (1 + log x). log x
= x^{x – 1} [1 + x log x . log ex]
∴ \(\frac{d y}{d x}\) = x^xx x^{x – 1} [1 + x log x log ex]
= x^{xx + x – 1} (1 + x log x log ex)

(iii) (sin x)^x = x^{sin x}
Answer:
Let y = u + v where u = (sin x)^x ………………… (1)
and v = x^{sin x} …………………… (2)

(iv) x^x + (cot x)^x
Answer:
Let y = x^x + (cot x)^x
and suppose y = u + v and
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………………… (1)
Where u = x^x …………………… (2)
and v = (cot x)^x …………………….. (3)
From (2)
u = x^x ⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = x^x (1 + log x)
From (3) v = (cot x)^x
⇒ log v = x log (cot x)
Differentiating both sides w.r.t to ‘x’

Question 2.
Establish the following (E.Q.)
(i) If x^y + y^x = a^b then
\(\frac{\mathbf{d y}}{\mathbf{d x}}\) = – \(\left(\frac{y \cdot x^{y-1}+y^x \cdot \log y}{x^y \log x+x \cdot y^{x-1}}\right)\)
Answer:
Let u = x^y ………………… (1) and v = y^x …………………… (2)
then given u + v = a^b ……………………… (3)
From (1), log u = y log x

(ii) If f(x) = sin^-1\(\sqrt{\frac{x-\beta}{\alpha-\beta}}\) and g(x) = tan^-1\(\sqrt{\frac{\mathbf{x}-\beta}{\alpha-\mathbf{x}}}\) then f'(x) = g'(x) (β < x < α) (March 2006)
Answer:

(iii) If a > b > 0 and 0 < x < π;
f(x) = (a² – b²)^-1/2 . cos^-1\(\left(\frac{a \cos x+b}{a+b \cos x}\right)\) then f'(x) = (a + b cos x)^-1
Answer:

Question 3.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) by
(i) Using product rule
(ii) Obtaining a single polynomial and expanding the product
(iii) Logarithmic differentiation.
Do they all give the same answer? (EQ.)
Answer:
(i) Using product rule:
\(\frac{d}{d x}\) (uv) = u\(\frac{d v}{d x}\) + v\(\frac{d u}{d x}\)
\(\frac{d}{d x}\)[(x² – 5x + 8) (x³ + 7x + 9)
= (x² – 5x + 8) \(\frac{d}{d x}\) (x³ + 7x + 9) + (x³ + 7x + 9) \(\frac{d}{d x}\) (x² – 5x + 8)
= (x² – 5x + 8) (3x² + 7) + (x² + 7x + 9) (2x – 5)
= 3x⁴ – 15x³ + 24x² + 7x² – 35x + 56 + 2x⁴ + 14x² + 18x – 5x³ – 35x – 45
= 5x⁴ – 20x³ + 45x² – 52x + 11

(ii) Expanding the product:
y = (x² – 5x + 8) (x³ + 7x + 9)
= x⁵ + 7x³ + 9x² – 5x⁴ – 35x² – 45x + 8x³ + 56x + 72
= x² – 5x⁴ + 15x³ – 26x² + 11x + 72
∴ \(\frac{d y}{d x}\) = 5x⁴ – 20x³ + 45x² – 52x + 11

(iii) By logarithmic differentiation:
y = (x² – 5x + 8) (x³ + 7x + 9)
log y = log (x² – 5x + 8) + log (x³ + 7x + 9)

= (2x – 5) (x³ + 7x + 9) + (3x² + 7) (x² – 5x + 8)
= 2x⁴ + 14x² + 18x – 5x³ – 35x – 45 + 3x⁴ – 15x³ + 24x² + 7x² – 35x + 56
= 5x⁴ – 20x³ + 45x² – 52x + 11
We observe that the same result is obtained by using the above three procedures.