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Answer: Hence Prove $\frac{dy}{dx}=\frac{4}{1+x^{2}}$

Hint:

\begin{aligned} &\frac{\mathrm{d}}{d \mathrm{x}}(\text { constan } \mathrm{t})=0 ; \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}

Given:

\begin{aligned} &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) \\ &0

Solution:

Prove :$\frac{dy}{dx}=\frac{4}{1+x^{2}}$

\begin{aligned} &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ &\text { Since } \sec ^{-1} x=\cos ^{-1}\left(\frac{1}{x}\right) \\ &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}

Let,

$x=\tan \theta$

$\theta =\tan ^{-1}x$

\begin{aligned} &y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\ &y=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right)+\cos ^{-1}(\cos 2 \theta) \end{aligned}

Using

\begin{aligned} &\sec ^{2} \theta=1+\tan ^{2} \theta \\ &\frac{1}{\sec \theta}=\cos \theta \\ &\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta \end{aligned}

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right)+\cos ^{-1}(\operatorname{Cos} 2 \theta) \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta)+\cos ^{-1}(\cos 20) \\ &\mathrm{y}=\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \theta)-(\mathrm{i}) \\ &\mathrm{u} \operatorname{sing} 2 \sin \theta \cos \theta=\sin 2 \theta \end{aligned}

Considering Limits,

\begin{aligned} &0<\mathrm{x}<1 \\ &0<\tan \theta<1 \\ &0<\theta<\frac{\mathrm{\pi}}{4} \\ &0<(2 \theta)<\frac{\mathrm{\pi}}{2} \end{aligned}

so from $e_{i}\: \: \left ( i \right )$

$y=2\theta +2\theta$

$y=4\theta$

\begin{aligned} &\cos ^{-1}(\cos \theta) \Rightarrow \text { if } \theta \in[0, \pi] \\ &y=4 \tan ^{-1} x \end{aligned}                                                    $since,x=\tan \theta ,\theta =\tan ^{-1}x$

Differentiating it with respect to x

\begin{aligned} &\frac{d y}{d x}=4 \frac{d}{d x}\left(\tan ^{-1} x\right) \\ &\text { Using } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=4\left(-\frac{1}{1+x^{2}}\right) \\ &\frac{d y}{d x}=\frac{4}{1+x^{2}} \end{aligned}

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