Chapter 1 - Differentiation
Exercise 1.2
Q.7 Differentiate the following w. r. t. x.
i)
$\begin{aligned}
y&=\cot ^{-1}\left[\cot \left(e^{x^{2}}\right)\right]\\
&Using \quad {cot }^{-1}(\cot x)=x\\
y &=e^{x^{2}} \\
&Diff \quad w.r.t \quad x \\
\frac{d y}{d x} &=\frac{d}{d x}\left(e^{x^{2}}\right) \\
&=e^{x^{2}} \cdot \frac{d}{d x} x^{2} \\
&=e^{x^{2}} \cdot 2 x \\
&=2 x e^{x^{2}}\\
\end{aligned}$
ii)
$\begin{aligned}
y&=\operatorname{cosec}^{-1}\left(\frac{1}{\cos 5^{x}}\right)\\
&=\operatorname{cosec}^{-1}\left(\sec 5^{x}\right) & \because \frac {1}{\cos x}=\sec x\\
&y=\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{\pi}{2}-5^{x}\right)\right] & \because \sec x=cosec(\pi/2-x)\\
&y=\frac{\pi}{2}-5^{x} & \because \operatorname{cosec}^{-1}(\operatorname{cosec}x) =x\\\\
&Diff \quad w.r.t. \quad x\\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{2}-5^{x}\right)\\
&=-5^{x} \cdot \log 5\end{aligned}$
iii)
$\begin{aligned}
y&=\cos ^{-1}(\sqrt{\frac{1+\cos x}{2}})\\\\
Using \\
[\cos 2 x&=2 \cos ^{2} x-1 \\
\cos 2 x+1&=2 \cos ^{2} x \\
\cos ^{2} x&=\frac{1+\cos 2 x}{2}\\
\frac{1+\cos x}{2}&=\cos ^{2}\left(\frac{x}{2}\right)]\\\\
&y=\cos ^{-1}(\sqrt{\cos ^{2}\left(\frac{x}{2}\right)}\\
&=\cos ^{-1}\left(\cos \frac{x}{2}\right)\\
&y=\frac{x}{2} \\
\frac{d y}{d x}&=\frac{x}{2} \\
&=\frac{1}{2}\frac{d}{d x} x\\
&=\frac{1}{2}
\end{aligned}$
iv)
$\begin{aligned}
y&=\cos ^{-1}(\sqrt{\frac{1-\cos \left(x^{2}\right)}{2}}\\
using \\
\cos 2 x&=1-2 \sin ^{2} x\\
2 \sin ^{2} x&=1-\cos 2 x\\
\sin ^{2} x&=\frac{1-\cos 2 x}{2}\\
\sin^{2}\left(\frac{x^{2}}{2}\right)&=\frac{1-\cos \left(x^{2}\right)}{2}\\\\
y&=\cos ^{-1} \sqrt{\sin ^{2}\left(\frac{x^{2}}{2}\right)}\\
y&=\cos ^{-1}\left(\sin \left(\frac{x^{2}}{2}\right)\right)\\
y&=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-\frac{x^{2}}{2}\right)\right) & \because \sin x=\cos(\pi/2-x)\\
y&=\frac{\pi}{2}-\frac{x^{2}}{2}\\
&Diff \quad w.r.t.\quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x^{2}}{2}\right)\\
&=0-\frac{1}{2} \cdot 2 x\\
\frac{d y}{d x}&=-x\end{aligned}$
v)
$\begin{aligned}
y&=\tan ^{-1}\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)\\\\
&using\\
&\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\\
&\tan 45^{\circ}=\tan \left(\frac{\pi}{4}\right)^{c}=1\\
&\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}=\frac{\tan \left(\frac{\pi}{4}\right)-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{\pi}{4}\right) \cdot \tan \left(\frac{x}{2}\right)}=\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\\\\
y&=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)\\
y&=\frac{\pi}{4}-\frac{x}{2}\\
&Diff \quad w.r.t \quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{4}-\frac{x}{2}\right)\\
&=0-\frac{1}{2} \cdot \frac{d}{d x}(x)\\
&=\frac{-1}{2}\end{aligned}$
vi)
$\begin{aligned}
y&=\operatorname{cosec}^{-1}\left(\frac{1}{4 \cos ^{3} 2 x-3 \cos 2 x}\right)\\
&using \\
&\cos 3 x =4 \cos ^{3} x-3 \cos x\\
y&=\operatorname{cosec}^{-1}\left(\frac{1}{\cos 3 \cdot(2 x)}\right)\\
&=\operatorname{cosec}^{-1}\left(\frac{1}{\cos 6 x}\right)\\
&=\operatorname{cosec}^{-1}(\sec 6 x)\\
&=\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{\pi}{2}-6 x\right)\right) \quad \because \sec x=\operatorname{cosec}\left(\frac{\pi}{2}-x\right)\\
y&=\frac{\pi}{2}-6 x\\
&Diff \quad wrt \quad x\\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{2}-6 x\right)\\
\frac{d y}{d x}&=-6
\end{aligned}$
vii)
$\begin{aligned}
y&=tan ^{-1}\left(\frac{1+\cos \left(\frac{x}{3}\right)}{\sin \frac{x}{3}}\right)\\
&Using\\
&\cos 2 x= 2\cos ^{2} x-1\\
&2 \cos ^{2} x=\cos 2 x+1\\
&2\cos ^{2} x ={1+\cos 2 x}\\
&2 \cdot \cos ^{2}\left(\frac{x}{6}\right)=1+\cos \left(\frac{x}{3}\right)\\
y&=\tan ^{-1}\left(\frac{2 \cdot \cos ^{2}\left(\frac{x}{6}\right)}{2 \cdot \sin \left(\frac{x}{6}\right) \cdot \cos \left(\frac{x}{6}\right)}\right) & \because \sin 2x =2 \sin x \cos x\\
y&=\tan ^{-1}\left(\frac{\cot (\frac{x}{6})}{\sin \left(\frac{x}{6}\right)}\right)\\
y&=\tan ^{-1}\left(\cot \left(\frac{x}{6}\right)\right)\\
y&=\tan ^{-1}\left(\tan ^{-1}\left(\frac{\pi}{2}-\frac{x}{6}\right)\right)\\
y&=\frac{\pi}{2}-\frac{x}{6}\\
&Diff \quad w.r.t \quad x \\
\frac{d y}{d x}&=-\frac{1}{6}\\
\end{aligned}$
viii)
$\begin{aligned}
y&=\cot ^{-1}\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\\
using \\
&\Rightarrow \cos 2 x=2 \cos ^{2} x-1\\
&\cos 2 x+1=2 \cos ^{2} x\\
&\cos 2\left(\frac{3}{2} x\right)+1=2 \cos ^{2}\left(\frac{3}{2}\right){x}\\
&\therefore \cos 3 x+1=2 \cos ^{2}\left(\frac{3}{2}\right) x\\
&\Rightarrow \sin 2 x=2 \sin x \cos x\\
&\therefore\sin 3 x=2 \sin \left(\frac{3}{2}\right) x \cdot \cos \left(\frac{3}{2}\right) x\\\\
Now,\\
y&=\cot ^{-1}\left(\frac{2 \sin \left(\frac{3 x}{2}\right) \cos \left(\frac{3 x}{2} x\right)}{2 \cdot \cos ^{2}\left(\frac{3}{2}\right) x}\right)\\
y&=\cot ^{-1}\left(\tan \left(\frac{3 x}{2}\right)\right)\\
y&=\cot ^{-1}\left(\cot \left(\frac{\pi}{2}-\frac{3}{2} x\right)\right) \quad \because \tan \theta=\cot \left(\frac{\pi}{2}-\theta\right)\\
y&=\frac{\pi}{2}-\frac{3}{2} x\\
&Diff \quad w.r.t \quad x\\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{3}{2} x\right)\\
&=-\frac{3}{2}\end{aligned}$
ix)
$\begin{aligned}
y&=\tan ^{1}\left(\frac{\cos 7 x}{1+\sin 7 x}\right)\\\\
&Using\\
&\cos 2x=cos^2x-sin^2x\\
&(cosx+sinx)^2=cos^2x+sin^2x+2sinxcosx =1+sin2x\\\\
y&=\tan^{-1}\left(\frac{\cos^{2}\left(\frac{7x}{2}\right)-\sin ^{2}\left(\frac{7x}{2}\right)}{\left[\cos\left(\frac{7 x}{2}\right)+\sin \left(\frac{7 x}{2}\right)\right]^{2}}\right] \\\\
y&=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{7 x}{2}\right)-\sin \left(\frac{7 x}{2}\right)\right]\left[\cos \left(\frac{7 x}{2}\right)+\sin \left(\frac{7 x}{2}\right)\right]}{\left[\cos \left(\frac{7 x}{2}\right)+\sin \left(\frac{7 x}{2}\right)\right]\left[\cos \left(\frac{7 x}{2}\right)+\sin \left(\frac{7 x}{2}\right)\right]}\right)\\\\
y&=\tan ^{-1}\left(\frac{\cos \left(\frac{7 x}{2}\right)-\sin \left(\frac{7 x}{2}\right)}{\cos \left(\frac{7 x}{2}\right)+\sin \left(\frac{7 x}{2}\right)}\right)\\
&Divide \quad by \quad \cos \left(\frac{7x}{2}\right)\\\\
y&=\tan ^{-1}\left(\frac{1-\tan \left(\frac{7x}{2}\right)}{1+\tan \left(\frac{7x}{2}\right)}\right)\\\\
y&=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}- \frac{7x}{2}\right)\right)\\
y&=\frac{\pi}{4}-\frac{7 x}{2}\\
&Diff \quad w.r.t \quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{4}-\frac{7 x}{2}\right)\\
&=\frac{-7}{2}\end{aligned}$
x)
$\begin{aligned}
y&=\tan ^{-1}(\sqrt{\frac{1+\cos x}{1-\cos x}})\\
&Using\\
&\Rightarrow \cos 2 x=2 \cos ^{2} x-1\\
&1+\cos 2 x=2 \cos ^{2} x\\
&\therefore 1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right)\\
&\Rightarrow \cos 2 x=1-2 \sin ^{2} x\\
&1-\cos 2 x=2 \sin ^{2} x\\
&\therefore1-\cos x=2 \sin ^{2} \frac{x}{2}\\
y&=\tan ^{-1}\left(\sqrt{\frac{2 \cdot \cos ^{2}\left(\frac{x}{2}\right)}{2 \cdot \sin ^{2}\left(\frac{x}{2}\right)}}\right)\\
y&=\tan ^{-1}\left(\sqrt{\cot ^{2} \frac{x}{2}}\right)\\
y&=\tan ^{-1}\left(\cot \frac{x}{2}\right)\\
y&=\tan ^{-1}\left(\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right)\\
y&=\frac{\pi}{2}-\frac{x}{2}\\
&Diff \quad w.r.t \quad x\\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{2}\right)\\
&=\frac{-1}{2}\end{aligned}$
xi)
$\begin{aligned}
y&=\tan ^{-1}(\operatorname{cosec} x+\cot x)\\
y&=\tan ^{-1}\left(\frac{1}{\sin x}+\frac{\cos x}{\sin x}\right)\\
&=\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\\
&Using\\
&\Rightarrow \cos 2 x=2 \cos ^{2} x-1\\
&2 \cos ^{2} x=1+\cos 2 x\\
&2 \cos ^{2}\left(\frac{x}{2}\right)=1+\cos x\\
&\Rightarrow \sin 2 x=2 \sin x \cdot \cos x\\
y&=\tan ^{-1}\left(\frac{2 \cdot \cos ^{2}\left(\frac{x}{2}\right)}{2 \cdot \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}\right) \\
y&=\tan ^{-1}\left(\frac{\cos \left(x_{2}\right)}{\sin \left(\frac{x}{2}\right)}\right) \\
y&=\tan ^{-1}\left(\cot \left(\frac{x}{2}\right)\right)\\
y&=\tan ^{-1}\left(\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right)\\
y&=\frac{\pi}{2}-\frac{x}{2}\\
& Diff \quad w.r.t. \quad x\\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{2}\right)\\
&=\frac{-1}{2}\end{aligned}$
xii)
$\begin{aligned}
y&=\cot ^{-1}\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \frac{4 x}{3}}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right.})}\right)\\\\
y&=\cot ^{-1}\left(\frac{\sqrt{\left[\cos \frac{2 x}{3}+\sin \frac{2 x}{3}\right]^{2}}+\sqrt{\left[\cos \frac{2 x}{3}-\sin \frac{2 x}{3}\right]^{2}}}{\sqrt{\left[\cos \frac{2 x}{3}+\sin \frac{2 x}{3}\right]^{2}}-\sqrt{\left[\cos \frac{2 x}{3}-\sin \frac{2 x}{3}\right]^{2}}}\right)\\\\
y&=\cot ^{-1}\left(\frac{\cos \left(\frac{2x}{3} \right)+\sin \left(\frac{2 x}{3}\right)+\cos \left(\frac{2 x}{3}\right)-\sin \left(\frac{2 x}{3}\right)}{\cos \left(\frac{2 x}{3}\right)+\sin \left(\frac{2 x}{3}\right)-\cos \left(\frac{2 x}{3}\right)+\sin \left(\frac{2 x}{3}\right)}\right)\\\\
y&=\cot ^{-1}\left(\frac{2 \cdot \cos \left(\frac{2 x}{3}\right)}{2 \sin \left(\frac{2 x}{3}\right)}\right)\\\\
y&=\cot ^{-1}\left(\frac{\left.\cos \frac{2 x}{3}\right)}{\sin \left(\frac{2 x}{3}\right)}\right)\\\\
y&=\cot ^{-1}\left(\cot \left(\frac{2 x}{3}\right)\right)\\
y&=\frac{2 x}{3}\\
&Diff \quad w.r.t \quad x\\
\frac{d y}{d x}&=\frac{2}{3}\end{aligned}$
Exercise 1.2
Tag
(7) Differentiate the following w. r. t. x.
(i) cot−1 [cot (ex2)]
(ii) cosec−1(1/cos (5x)
(iii) cos−1 sqrt((1 + cos x)/2)
(iv) cos−1 sqrt((1 − cos (x^2)/2)
(v) tan−1(1 − tan (x/2))/(1 + tan (x/2))
(vi) cosec^−1(1/(4 cos^32x − 3 cos 2x))
(vii) tan−1(1 + cos (x/3))/sin (x/3))
(viii) cot−1(sin 3x/(1 + cos 3x))
(ix) tan−1t(cos 7x/(1 + sin 7x))
(x) tan−1sqrt((1 + cos x)/(1 − cos x))
(xi) tan−1sqrt(cosec x + cot x)
(xii) cot−1(1 + sin (4x3 ) + 1 − sin (4x3 )1 + sin (4x3 ) − 1 − sin (4x3 ))
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