Chapter 1 - Differentiation
Exercise 1.4
Q.1 (1) Find dy/dx if
i)
$\begin{aligned}
x&=a t, y=2 a t\\
Given, \\
x&=a t^{2}\\
&Diff\quad w.r.t. \quad t \\
\frac{d x}{d t}&=\frac{d}{d t}a t^{2}\\
\frac{d x}{d t}&=2 a t\\\\
And,\\
&y=2 a t\\
&Diff \quad w.r.t. \quad t \\
\frac{d y}{d t}&=2 a \cdot \frac{d}{d t}(t)\\
&=2a\\
Now,\\
\frac{d y}{d x}&=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 a}{2 a t}\\
&=\frac{1}{t}\end{aligned}$
ii)
$\begin{aligned}
x&=a \cot \theta, y=b \operatorname{cosec} \theta\\
Given,\\
x&=a \cot \theta\\
&Diff \quad w.r.t. \quad \theta \\
\frac{d x}{d \theta} &=a \cdot \frac{d}{d \theta} \cot \theta \\
&=-a.cosec ^{2} \theta\\
And,\\
y&=b \operatorname{cosec} \theta\\
&Diff \quad w.r.t. \quad \theta \\
\frac{d y}{d \theta} &=b \cdot \frac{d}{d \theta} cosec \theta \\
&=-b \operatorname{cosec} \cot \theta\\
Now,\\
\frac{d y}{d x}&=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-b \operatorname{cosec} \cdot \cot \theta}{-a \operatorname{cosec}^{2} \theta}\\
&=\frac{b}{a} \frac{\cot \theta}{\operatorname{cosec} \theta}\\
&=\frac{b}{a} \frac{\frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta}}\\
&=\frac{b}{a} \cos \theta \end{aligned}$
iii)
$\begin{aligned}
x& = \sqrt{a^{2}+m^{2}} \quad, \quad y=\log \left(a^{2}+m^{2}\right)\\
Given,\\
x& =\sqrt{a^{2}+m^{2}}\\
& Diff \quad w.r.t. \quad m \\
\frac{d x}{d m}&=\frac{d}{d m}(\sqrt{a^{2}+m^{2}})\\
& =\frac{1}{2 \sqrt{a^{2}+m^{2}}} \frac{d}{d m}\left(a^{2}+m^{2}\right)\\
& =\frac{1}{2 \sqrt{a^{2}+m^{2}}} \cdot 2 m\\
& =\frac{1}{\sqrt{a^{2}+m^{2}}} \cdot m\\
And,\\
y& =\log \left(a^{2}+m^{2}\right) \\
& Diff \quad w.r.t. \quad m \\
\frac{d y}{d m}& =\frac{d}{d m} \log \left(a^{2}+m^{2}\right)\\
& =\frac{1}{a^{2}+m^{2}} \cdot \frac{d}{d m}\left(a^{2}+m^{2}\right)\\
\frac{d y}{d m}& =\frac{1}{a^{2}+m^{2}} \cdot 2 m\\
Now,\\
\frac{d y}{d x}& =\frac{\frac{d y}{d m}}{\frac{d x}{d m}}\\\\
& =\frac{\frac{2 m}{a^{2}+m^{2}}} {\frac{m}{\sqrt{a^{2}+m^{2}}}}\\ \\
&=\frac{2}{\sqrt{a^{2}+m^{2}}} \end{aligned}$
iv)
$\begin{aligned}
x&=\sin \theta,y=\tan \theta\\
Give,\\
x&=\sin \theta\\
&Diff \quad w.r.t \quad \theta\\
\frac{d x}{d \theta}&=\frac{d}{d \theta} \sin \theta\\
&=\cos \theta\\
And,\\
y&=\tan \theta\\
&Diff \quad w.r.t \quad \theta\\
\frac{d y}{d \theta}&=\frac{d }{d \theta}(\tan \theta)\\
&=\sec ^{2} \theta\\
Now,\\
\frac{d y}{d x}&=\frac{d y}{d \theta} \times \frac{d \theta}{d x} \\
or\\
\frac{d y}{d x}&=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\
&=\frac{\sec ^{2} \theta}{\cos \theta}\\
&=\frac{1}{\cos ^{2} \theta \cdot \cos \theta} \\\\
&=\frac{1}{\cos ^{3} \theta}\\\\
&=\sec ^{3} \theta\\
\end{aligned}$
v)
$\begin{aligned}\\
x&=a(1-\cos \theta), y=b(\theta-\sin \theta)\\
Given\\
x&=a(1-\cos \theta)\\
& Diff \quad w.r.t.\quad \theta\\
\frac{d x}{d \theta}&=a \cdot \frac{d}{d \theta}(1-\cos \theta)\\
&=a \sin \theta\\
And,\\
y&=b(\theta-\sin \theta)\\
& Diff \quad w.r.t.\quad \theta\\
\frac{d y}{d \theta}&=b \cdot \frac{d}{d \theta}(\theta-\sin \theta)\\
&=b(1-\cos \theta)\\
Now,\\
\frac{d y}{d x}&=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\\\\
&=\frac{b(1-\cos \theta)}{a \sin \theta}\\
&=\frac{b}{a} \frac{2 \cdot \sin ^{2} \left(\frac{\theta}{2}\right)}{2 \cdot \sin \left(\frac{\theta}{2}\right)\cos \left(\frac{\theta}{2}\right)} \\
&=\frac{b}{a} \cdot \frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)}\\
&=\frac{b}{a} \tan \left(\frac{\theta}{2}\right)
\end{aligned}$
vi)
$\begin{aligned}
x&=\left(t+\frac{1}{t}\right)^{a}, y=a^{t+\frac{1}{t}}\\
Given,\\
&put, \left(t+\frac{1}{t}\right)=m\\
x&=m^{a}\\
&Diff \quad w.r.t. \quad m \\
\frac{d x}{d m}&=\frac{d}{d m}(m)^{a}\\
&=a.m^{a-1}\\
And,\\
y&=a^{t+\frac{1}{t}} \\ y &=a^{m}\\
&Diff \quad w.r.t. \quad m \\
\frac{d y}{d m}&=\frac{d}{d m}\left(a^{m}\right)\\
&=a^{m} \cdot \log a\\
Now,\\
\frac{d y}{d x}&=\frac{\frac{d y}{d x}}{\frac{d x}{d x}}\\\\
&=\frac{a^{m} \log a}{a \cdot m^{a-1}}\\\\
&=\frac{a^{m} \log a}{a \cdot m^{a}.m^{-1}}\\\\
&=\frac{a^{m} \log a \cdot m}{a \cdot m^{a}}\\\\
&=\frac{y}{a x} \log a\left(1+\frac{1}{t}\right)\\\\
&=\frac{y}{x a} \log a\left(\frac{t^{2}+1}{t}\right)\\\\
&=\frac{4\left(t^{2}+1\right) \log a}{x a}\end{aligned}$
vii)
$\begin{aligned}
x&=\cos ^{2}\left(\frac{2 t}{1+t^{2}}\right) ; y=\sec ^{-1}(\sqrt{1+t^{2}})\\
Given,\\
x&=\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\\
put, \quad t&=\tan \theta\\
x&=\cos ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\\
x&=\cos ^{-1}( \sin 2 \theta)\\
x&=\cos ^{-1}\left(\cos ^{-1}\left(\frac{\pi}{2}-2 \theta\right)\right)\\
x&=\frac{\pi}{2}- 2 \theta\\
&Diff \quad w.r.t \quad \theta\\
\frac{d x}{d \theta}&=\frac{d}{d \theta}\left(\frac{\pi}{2}-2\theta \right)\\
&=-2\\
And\\
y&=\sec ^{-1}(\sqrt{1+t^{2}})\\
& put\quad t=\tan \theta\\
y&=\sec ^{-1}(\sqrt{1+\tan ^{2} \theta})\\
y&=\sec ^{-1}(\sqrt{\sec ^{2} \theta})\\
y&=\sec ^{2}(\sec \theta)\\
y&=\theta\\
&Diff \quad w.r.t \quad \theta\\
\frac{d y}{d \theta}&=1\\\\
Now,\\
\frac{d y}{d x}&=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\\
&=\frac{1}{-2}\end{aligned}$
viii)
$\begin{aligned}
x&=\cos ^{1}\left(4 t^{3}-3 t\right), y=\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\\
Given,\\
x&=\cos ^{-1}\left(4 t^{3}-3 t\right)\\
&Put,\quad t=\cos \theta \\
x &=\cos ^{-1}\left(4 \cos ^{3} \theta-2 \cos \theta\right) \\
x &=\cos ^{-1}(\cos 3 \theta) \\
x &=3 \theta \\
& Diff \quad w.r.t. \quad \theta \\
\frac{d x}{d \theta}&=3 \cdot \frac{d}{d \theta}(\theta)\\
&=3\\
And,\\
y&=\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\\
&Put,\quad t=\cos \theta \\
y&=\tan ^{-1}\left(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\right)\\
&=\tan ^{-1}\left(\frac{\sqrt{\sin ^{2} \theta}}{\cos\theta}\right)\\
&=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)\\
&=\tan ^{-1}(\tan \theta) \\
y&=\theta\\
& Diff \quad w.r.t. \quad \theta \\
\frac{d y}{d \theta}&=1\\\\
Now,\\
\frac{d y}{d x}&=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\\
&=\frac{1}{3}\end{aligned}$
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