# Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.4 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

Question 1.
Find $$\frac{d y}{d x}$$ if
(i) x = at2, y = 2at
Solution:
x = at2, y = 2at
Differentiating x and y w.r.t. t, we get

(ii) x = a cot θ, y = b cosec θ
Solution:
x = a cot θ, y = b cosec θ
Differentiating x and y w.r.t. θ, we get

(iii) x = $$\sqrt{a^{2}+m^{2}}$$, y = log (a2 + m2)
Solution:
x = $$\sqrt{a^{2}+m^{2}}$$, y = log (a2 + m2)
Differentiating x and y w.r.t. m, we get
$$\frac{d x}{d m}=\frac{d}{d m}\left(\sqrt{a^{2}+m^{2}}\right)$$

(iv) x = sin θ, y = tan θ
Solution:
x = sin θ, y = tan θ
Differentiating x and y w.r.t. θ, we get

(v) x = a(1 – cos θ), y = b(θ – sin θ)
Solution:
x = a(1 – cos θ), y = b(θ – sin θ)
Differentiating x and y w.r.t. θ, we get

(vi) x = $$\left(t+\frac{1}{t}\right)^{a}$$, y = $$a^{t+\frac{1}{t}}$$, where a > 0, a ≠ 1 and t ≠ 0
Solution:
x = $$\left(t+\frac{1}{t}\right)^{a}$$, y = $$a^{t+\frac{1}{t}}$$ ………(1)
Differentiating x and y w.r.t. t, we get

(vii) x = $$\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$$, y = $$\sec ^{-1}\left(\sqrt{1+t^{2}}\right)$$
Solution:
x = $$\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$$, y = $$\sec ^{-1}\left(\sqrt{1+t^{2}}\right)$$
Put t = tan θ Then θ = tan-1t

(viii) x = cos-1(4t3 – 3t), y = $$\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)$$
Solution:
x = cos-1(4t3 – 3t), y = $$\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)$$
Put t = cos θ. Then θ = cos-1t
x = cos-1(4cos3θ – 3cos θ)

Question 2.
Find $$\frac{d y}{d x}$$, if
(i) x = cosec2θ, y = cot3θ at θ = $$\frac{\pi}{6}$$
Solution:
x = cosec2θ, y = cot3θ
Differentiating x and y w.r.t. θ, we get

(ii) x = a cos3θ, y = a sin3θ at θ = $$\frac{\pi}{3}$$
Solution:
x = a cos3θ, y = a sin3θ
Differentiating x and y w.r.t. θ, we get

(iii) x = t2 + t + 1, y = sin($$\frac{\pi t}{2}$$) + cos($$\frac{\pi t}{2}$$) at t = 1
Solution:
x = t2 + t + 1, y = sin($$\frac{\pi t}{2}$$) + cos($$\frac{\pi t}{2}$$)
Differentiating x and y w.r.t. t, we get

(iv) x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = $$\frac{\pi}{4}$$
Solution:
x = 2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get

(v) x = t + 2 sin(πt), y = 3t – cos(πt) at t = $$\frac{1}{2}$$
Solution:
x = t + 2 sin(πt), y = 3t – cos(πt)
Differentiating x and y w.r.t. t, we get

Question 3.
(i) If x = $$a \sqrt{\sec \theta-\tan \theta}$$, y = $$a \sqrt{\sec \theta+\tan \theta}$$, then show that $$\frac{d y}{d x}=-\frac{y}{x}$$
Solution:
x = $$a \sqrt{\sec \theta-\tan \theta}$$, y = $$a \sqrt{\sec \theta+\tan \theta}$$

(ii) If x = $$e^{\sin 3 t}$$, y = $$e^{\cos 3 t}$$, then show that $$\frac{d y}{d x}=-\frac{y \log x}{x \log y}$$
Solution:
x = $$e^{\sin 3 t}$$, y = $$e^{\cos 3 t}$$
log x = log $$e^{\sin 3 t}$$, log y = log $$e^{\cos 3 t}$$
log x = (sin 3t)(log e), log y = (cos 3t)(log e)
log x = sin 3t, log y = cos 3t ….. (1) [∵ log e = 1]
Differentiating both sides w.r.t. t, we get

(iii) If x = $$\frac{t+1}{t-1}$$, y = $$\frac{1-t}{t+1}$$, then show that y2 – $$\frac{d y}{d x}$$ = 0.
Solution:
x = $$\frac{t+1}{t-1}$$, y = $$\frac{1-t}{t+1}$$

(iv) If x = a cos3t, y = a sin3t, then show that $$\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$$
Solution:
x = a cos3t, y = a sin3t
Differentiating x and y w.r.t. t, we get

(v) If x = 2 cos4(t + 3), y = 3 sin4(t + 3), show that $$\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}$$
Solution:
x = 2 cos4(t + 3), y = 3 sin4(t + 3)

(vi) If x = log (1 + t2), y = t – tan-1t, show that $$\frac{d y}{d x}=\frac{\sqrt{e^{x}-1}}{2}$$
Solution:
x = log (1 + t2), y = t – tan-1t
Differentiating x and y w.r.t. t, we get

(vii) If x = $$\sin ^{-1}\left(e^{t}\right)$$, y = $$\sqrt{1-e^{2 t}}$$, show that sin x + $$\frac{d y}{d x}$$ = 0
Solution:
x = $$\sin ^{-1}\left(e^{t}\right)$$, y = $$\sqrt{1-e^{2 t}}$$
Differentiating x and y w.r.t. t, we get

(viii) If x = $$\frac{2 b t}{1+t^{2}}$$, y = $$a\left(\frac{1-t^{2}}{1+t^{2}}\right)$$, show that $$\frac{d x}{d y}=-\frac{b^{2} y}{a^{2} x}$$
Solution:
x = $$\frac{2 b t}{1+t^{2}}$$, y = $$a\left(\frac{1-t^{2}}{1+t^{2}}\right)$$

Question 4.
(i) Differentiate x sin x w.r.t tan x.
Solution:
Let u = x sinx and v = tan x
Then we want to find $$\frac{d u}{d v}$$
Differentiating u and v w.r.t. x, we get

(ii) Differentiate $$\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$$ w.r.t $$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$$
Solution:
Let u = $$\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$$ and v = $$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$$
Then we want to find $$\frac{d u}{d v}$$

(iii) Differentiate $$\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$$ w.r.t $$\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)$$
Solution:

(iv) Differentiate $$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$$ w.r.t. tan-1x
Solution:
Let u = $$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$$ and v = tan-1x
Then we want to find $$\frac{d u}{d v}$$
Put x = tan θ. Then θ = tan-1x.

(v) Differentiate 3x w.r.t. logx3.
Solution:
Let u = 3x and v = logx3.
Then we want to find $$\frac{d u}{d v}$$
Differentiating u and v w.r.t. x, we get
$$\frac{d u}{d x}=\frac{d}{d x}\left(3^{x}\right)=3^{x} \cdot \log 3$$

(vi) Differentiate $$\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$$ w.r.t. sec-1x.
Solution:
Let u = $$\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$$ and v = sec-1x
Then we want to find $$\frac{d u}{d v}$$.
Differentiating u and v w.r.t. x, we get

(vii) Differentiate xx w.r.t. xsin x.
Solution:
Let u = xx and v = xsin x
Then we want to find $$\frac{d u}{d x}$$.
Take, u = xx
log u = log xx = x log x
Differentiating both sides w.r.t. x, we get

(viii) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$$
Solution:
Let u = $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ and v = $$\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$$
Then we want to find $$\frac{d u}{d v}$$
u = $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$
Put x = tan θ. Then θ = tan-1x and