# Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.3

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

## Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.3

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 $\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}$
= 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

(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 $\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}$

Differentiating both sides w.r.t. x, we get

(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 $\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}$

Differentiating both sides w.r.t. x, we get

(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 $\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}$

Differentiating both sides w.r.t. x, we get

(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 $\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}$
= log x5 + log tan34x – log sin23x
= 5 log x+ 3 log (tan 4x) – 2 log (sin 3x)
Differentiating both sides w.r.t. x, we get

(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

(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

(viii) sin xx
Solution:
Let y = (sin xx)
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 = xx
Then log u = log xx = x . log x
Differentiating both sides w.r.t. x, we get

Question 2.
Differentiate the following w.r.t. x:
(i) xe + xx + ex + ee
Solution:
Let y = xe + xx + ex + ee
Let u = xx
Then log u = log xx = x log x
Differentiating both sides w.r.t. x, we get
$$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x \log x)$$

(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 $$x^{x^{x}}$$ = xx . log x
Differentiating both sides w.r.t. x, we get

(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

(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 $$x^{e^{x}}$$ = ex . log x
Differentiating both sides w.r.t. x, we get

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
$$\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}[(\sin x) \cdot(\log \log x)]$$
= $$(\sin x) \cdot \frac{d}{d x}\left[(\log \log x)+(\log \log x) \cdot \frac{d}{d x}(\sin x)\right]$$

(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

(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

(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)

(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 $\left[(\tan x)^{\tan x}\right]^{\tan x}$
= 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

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

(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

(iii) x + √xy + y = 1
Solution:
x + √xy + y = 1
Differentiating both sides w.r.t. x, we get

(iv) x3 + x2y + xy2 + y3 = 81
Solution:
x3 + x2y + xy2 + y3 = 81
Differentiating both sides w.r.t. x, we get

(v) x2y2 – tan-1($$\sqrt{x^{2}+y^{2}}$$) = cot-1($$\sqrt{x^{2}+y^{2}}$$)
Solution:
x2y2 – tan-1($$\sqrt{x^{2}+y^{2}}$$) = cot-1($$\sqrt{x^{2}+y^{2}}$$)
∴ x2y2 = tan-1($$\sqrt{x^{2}+y^{2}}$$) + cot-1($$\sqrt{x^{2}+y^{2}}$$)
∴ x2y2 = $$\frac{\pi}{2}$$ …….[∵ $$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$$]
Differentiating both sides w.r.t. x, we get

(vi) xey + yex = 1
Solution:
xey + yex = 1
Differentiating both sides w.r.t. x, we get

(vii) ex+y = cos (x – y)
Solution:
ex+y = cos (x – y)
Differentiating both sides w.r.t. x, we get

(viii) cos (xy) = x + y
Solution:
cos (xy) = x + y
Differentiating both sides w.r.t. x, we get

(ix) $$e^{e^{x-y}}=\frac{x}{y}$$
Solution:
$$e^{e^{x-y}}=\frac{x}{y}$$
∴ ex-y = log($$\frac{x}{y}$$) …….[ex = y ⇒ x = log y]
∴ ex-y = log x – log y
Differentiating both sides w.r.t. x, we get

Question 4.
Show that $$\frac{d y}{d x}=\frac{y}{x}$$ in the following, where a and p are constants.
(i) x7y5 = (x + y)12
Solution:
x7y5 = (x + y)12
(log x7y5) = log(x + y)12
log x7 + log y5 = log(x + y)12
7 log x + 5 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get

(ii) xpy4 = (x + y)p+4, p∈N
Solution:
xpy4 = (x + y)p+4
Taking log
log (xpy4) = log(x + y)p+4
log xp + log y4 = (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

(iii) $$\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}$$
Solution:
$$\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}$$

Differentiating both sides w.r.t. x, we get

Differentiating both sides w.r.t. x, we get

(iv) $$\tan ^{-1}\left(\frac{3 x^{2}-4 y^{2}}{3 x^{2}+4 y^{2}}\right)=a^{2}$$
Solution:
$$\tan ^{-1}\left(\frac{3 x^{2}-4 y^{2}}{3 x^{2}+4 y^{2}}\right)=a^{2}$$

Differentiating both sides w.r.t. x, we get

(v) $$\cos ^{-1}\left(\frac{7 x^{4}+5 y^{4}}{7 x^{4}-5 y^{4}}\right)=\tan ^{-1} a$$
Solution:

(vi) $$\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20$$
Solution:
$$\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20$$

Differentiating both sides w.r.t. x, we get

(vii) $$e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a$$
Solution:
$$e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a$$

Differentiating both sides w.r.t. x, we get

(viii) $$\sin \left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=a^{3}$$
Solution:

Differentiating both sides w.r.t. x, we get

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

(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:
$$\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$$

(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:

Differentiating both sides w.r.t. x, we get

(iv) If ex + ey = ex+y, then show that $$\frac{d y}{d x}=-e^{y-x}$$
Solution:
ex + ey = ex+y ……(1)
Differentiating both sides w.r.t. x, we get

(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:
$$\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}$$
$$\frac{x^{5}-y^{5}}{x^{5}+y^{5}}=\sin \frac{\pi}{6}=\frac{1}{2}$$
2x5 – 2y5 = x5 + y5
3y5 = x5
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 xy = ex-y, then show that $$\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}}$$
Solution:
xy = ex-y
log xy = log ex-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 = $$\frac{x}{1+\log x}$$

(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:
$$y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}$$
y2 = cos x + $$\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}$$
y2 = cos x + y
Differentiating both sides w.r.t. x, we get

(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:
$$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}$$

(ix) If $$y=x^{x^{x^{-\infty}}}$$, then show that $$\frac{d y}{d x}=\frac{y^{2}}{x(1-\log y)}$$
Solution:
$$y=x^{x^{x^{-\infty}}}$$

(x) If ey = yx, then show that $$\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}$$
Solution:
ey = yx
log ey = log yx
y log e = x log y
y = x log y …… [∵log e = 1] ……….(1)
Differentiating both sides w.r.t. x, we get
$$\frac{d y}{d x}=x \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)$$