Differentiation Class 12 Maths 2 Exercise 1.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.3 Questions and Answers.

12th Maths Part 2 Differentiation Exercise 1.3 Questions And Answers Maharashtra Board

Question 1.
Differentiate the following w.r.t. x:
(i) \(\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}\)
Solution:
Let y = \(\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}\)
Then, log y = log [latex]\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}[/latex]
= log (x + 1) 2 – log (x + 2) 3 – log (x + 3) 4
= 2 log (x +1) – 3 log (x + 2) – 4 log (x + 3)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-i

(ii) \(\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}\)
Solution:
Let y = \(\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}\)
Then log y = log [latex]\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}[/latex]
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-ii
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-ii.1

Maharashtra-Board-Solutions

(iii) \(\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}\)
Solution:
Let y = \(\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}\)
Then log y = log [latex]\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}[/latex]
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-iii
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-iii.1

(iv) \(\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}\)
Solution:
Let y = \(\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}\)
Then log y = log [latex]\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}[/latex]
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-iv
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-iv.1
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-iv.2

(v) \(\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}\)
Solution:
Let y = \(\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}\)
Then log y = log [latex]\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}[/latex]
= log x 5 + log tan 3 4x – log sin 2 3x
= 5 log x+ 3 log (tan 4x) – 2 log (sin 3x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-v
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-v.1

(vi) \(x^{\tan ^{-1} x}\)
Solution:
Let y = \(x^{\tan ^{-1} x}\)
Then log y = log (\(x^{\tan ^{-1} x}\)) = (tan -1 x)(log x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-vi

(vii) (sin x) x
Solution:
Let y = (sin x) x
Then log y = log (sin x) x = x . log (sin x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-vii

(viii) sin x x
Solution:
Let y = (sin x x )
Then \(\frac{d y}{d x}=\frac{d}{d x}\left[\left(\sin x^{x}\right)\right]\)
\(\frac{d y}{d x}=\cos \left(x^{x}\right) \cdot \frac{d}{d x}\left(x^{x}\right)\) ……. (1)
Let u = x x
Then log u = log x x = x . log x
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q1-viii

Maharashtra-Board-Solutions

Question 2.
Differentiate the following w.r.t. x:
(i) x e + x x + e x + e e
Solution:
Let y = x e + x x + e x + e e
Let u = x x
Then log u = log x x = x log x
Differentiating both sides w.r.t. x, we get
\(\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x \log x)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-i

(ii) \(x^{x^{x}}+e^{x^{x}}\)
Solution:
Let y = \(x^{x^{x}}+e^{x^{x}}\)
Put u = \(x^{x^{x}}\) and v = \(e^{x^{x}}\)
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Take u = \(x^{x^{x}}\)
log u = log \(x^{x^{x}}\) = x x . log x
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-ii
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-ii.1
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-ii.2

(iii) (log x) x – (cos x) cot x
Solution:
Let y = (log x) x – (cos x) cot x
Put u = (log x) x and v = (cos x) cot x
Then y = u – v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\) ……..(1)
Take u = (log x) x
∴ log u = log (log x) x = x . log (log x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-iii
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-iii.1

(iv) \(x^{e^{x}}+(\log x)^{\sin x}\)
Solution:
Let y = \(x^{e^{x}}+(\log x)^{\sin x}\)
Put u = \(x^{e^{x}}\) and v = (log x) sin x
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……….(1)
Take u = \(x^{e^{x}}\)
∴ log u = log \(x^{e^{x}}\) = e x . log x
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-iv
Also, v = (log x) sin x
∴ log v = log (log x) sin x = (sin x) . (log log x)
Differentiating both sides w.r.t. x, we get
\(\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}[(\sin x) \cdot(\log \log x)]\)
= \((\sin x) \cdot \frac{d}{d x}\left[(\log \log x)+(\log \log x) \cdot \frac{d}{d x}(\sin x)\right]\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-iv.1

Maharashtra-Board-Solutions

(v) \(e^{\tan x}+(\log x)^{\tan x}\)
Solution:
Let y = \(e^{\tan x}+(\log x)^{\tan x}\)
Put u = (log x) tan x
∴ log u =log(log x) tan x = (tan x).(log log x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-v
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-v.1

(vi) (sin x) tan x + (cos x) cot x
Solution:
Let y = (sin x) tan x + (cos x) cot x
Put u = (sin x) tan x and v = (cos x) cot x
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………(1)
Take u = (sin x) tan x
∴ log u = log (sin x) tan x = (tan x) . (log sin x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-vi
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-vi.1

(vii) \(10^{x^{x}}+x^{x^{10}}+x^{10^{x}}\)
Solution:
Let y = \(10^{x^{x}}+x^{x^{10}}+x^{10^{x}}\)
Put u = \(10^{x^{x}}\), v = \(x^{x^{10}}\) and w = \(x^{10^{x}}\)
Then y = u + v + w
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}\) ………(1)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-vii
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-vii.1
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-vii.2
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-vii.3

(viii) \(\left[(\tan x)^{\tan x}\right]^{\tan x}\) at x = \(\frac{\pi}{4}\)
Solution:
Let y = \(\left[(\tan x)^{\tan x}\right]^{\tan x}\)
∴ log y = log [latex]\left[(\tan x)^{\tan x}\right]^{\tan x}[/latex]
= tan x . log(tan x) tan x
= tan x . tan x log (tan x)
= (tan x) 2 . log (tan x)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q2-viii

Question 3.
Find \(\frac{d y}{d x}\) if
(i) √x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-i

(ii) x√x + y√y = a√a
Solution:
x√x + y√y = a√a
∴ \(x^{\frac{3}{2}}+y^{\frac{3}{2}}=a^{\frac{3}{2}}\)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-ii

(iii) x + √xy + y = 1
Solution:
x + √xy + y = 1
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-iii

Maharashtra-Board-Solutions

(iv) x 3 + x 2 y + xy 2 + y 3 = 81
Solution:
x 3 + x 2 y + xy 2 + y 3 = 81
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-iv

(v) x 2 y 2 – tan -1 (\(\sqrt{x^{2}+y^{2}}\)) = cot -1 (\(\sqrt{x^{2}+y^{2}}\))
Solution:
x 2 y 2 – tan -1 (\(\sqrt{x^{2}+y^{2}}\)) = cot -1 (\(\sqrt{x^{2}+y^{2}}\))
∴ x 2 y 2 = tan -1 (\(\sqrt{x^{2}+y^{2}}\)) + cot -1 (\(\sqrt{x^{2}+y^{2}}\))
∴ x 2 y 2 = \(\frac{\pi}{2}\) …….[∵ \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)]
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-v

(vi) xe y + ye x = 1
Solution:
xe y + ye x = 1
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-vi

(vii) e x+y = cos (x – y)
Solution:
e x+y = cos (x – y)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-vii

(viii) cos (xy) = x + y
Solution:
cos (xy) = x + y
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-viii

(ix) \(e^{e^{x-y}}=\frac{x}{y}\)
Solution:
\(e^{e^{x-y}}=\frac{x}{y}\)
∴ e x-y = log(\(\frac{x}{y}\)) …….[e x = y ⇒ x = log y]
∴ e x-y = log x – log y
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q3-ix

Question 4.
Show that \(\frac{d y}{d x}=\frac{y}{x}\) in the following, where a and p are constants.
(i) x 7 y 5 = (x + y) 12
Solution:
x 7 y 5 = (x + y) 12
(log x 7 y 5 ) = log(x + y) 12
log x 7 + log y 5 = log(x + y) 12
7 log x + 5 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-i

Maharashtra-Board-Solutions

(ii) x p y 4 = (x + y) p+4 , p∈N
Solution:
x p y 4 = (x + y) p+4
Taking log
log (x p y 4 ) = log(x + y) p+4
log x p + log y 4 = (p + 4) log(x + y)
p log x + 4 log y = (p + 4) log(x + y)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-ii

(iii) \(\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}\)
Solution:
\(\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-iii
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-iii.1
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-iii.2
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-iii.3

(iv) \(\tan ^{-1}\left(\frac{3 x^{2}-4 y^{2}}{3 x^{2}+4 y^{2}}\right)=a^{2}\)
Solution:
\(\tan ^{-1}\left(\frac{3 x^{2}-4 y^{2}}{3 x^{2}+4 y^{2}}\right)=a^{2}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-iv
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-iv.1

(v) \(\cos ^{-1}\left(\frac{7 x^{4}+5 y^{4}}{7 x^{4}-5 y^{4}}\right)=\tan ^{-1} a\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-v
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-v.1

(vi) \(\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20\)
Solution:
\(\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-vi
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-vi.1

Maharashtra-Board-Solutions

(vii) \(e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a\)
Solution:
\(e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-vii
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-vii.1

(viii) \(\sin \left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=a^{3}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-viii
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q4-viii.1

Question 5.
(i) If log (x + y) = log (xy) + p, where p is a constant, then prove that \(\frac{d y}{d x}=-\frac{y^{2}}{x^{2}}\).
Solution:
log (x + y) = log (xy) + p
∴ log (x + y) = log x + log y + p
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-i

(ii) If \(\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2\), show that \(\frac{d y}{d x}=-\frac{99 x^{2}}{101 y^{2}}\)
Solution:
\(\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-ii

(iii) If \(\log _{5}\left(\frac{x^{4}+y^{4}}{x^{4}-y^{4}}\right)=2\), show that \(\frac{d y}{d x}=-\frac{12 x^{3}}{13 y^{3}}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-iii
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-iii.1

(iv) If e x + e y = e x+y , then show that \(\frac{d y}{d x}=-e^{y-x}\)
Solution:
e x + e y = e x+y ……(1)
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-iv

Maharashtra-Board-Solutions

(v) If \(\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}\), show that \(\frac{d y}{d x}=\frac{x^{4}}{3 y^{4}}\)
Solution:
\(\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}\)
\(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}=\sin \frac{\pi}{6}=\frac{1}{2}\)
2x 5 – 2y 5 = x 5 + y 5
3y 5 = x 5
Differentiating both sides w.r.t. x, we get
\(3 \times 5 y^{4} \frac{d y}{d x}=5 x^{4}\)
∴ \(\frac{d y}{d x}=\frac{x^{4}}{3 y^{4}}\)

(vi) If x y = e x-y , then show that \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}}\)
Solution:
x y = e x-y
log x y = log e x-y
y log x = (x – y) log e
y log x = (x – y) ….. [∵ log e = 1]
y + y log x = x – y
y + y log x = x
y(1 + log x) = x
y = \(\frac{x}{1+\log x}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-vi

(vii) If \(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}\), then show that \(\frac{d y}{d x}=\frac{\sin x}{1-2 y}\)
Solution:
\(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}\)
y 2 = cos x + \(\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}\)
y 2 = cos x + y
Differentiating both sides w.r.t. x, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-vii

(viii) If \(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}\), then show that \(\frac{d y}{d x}=\frac{1}{x(2 y-1)}\)
Solution:
\(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-viii

(ix) If \(y=x^{x^{x^{-\infty}}}\), then show that \(\frac{d y}{d x}=\frac{y^{2}}{x(1-\log y)}\)
Solution:
\(y=x^{x^{x^{-\infty}}}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-ix

Maharashtra-Board-Solutions

(x) If e y = y x , then show that \(\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}\)
Solution:
e y = y x
log e y = log y x
y log e = x log y
y = x log y …… [∵log e = 1] ……….(1)
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=x \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-x
Maharashtra-Board-12th-Maths-Solutions-Chapter-1-Differentiation-Ex-1.3-Q5-x.1