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Answer:$\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^{2}}}$

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}

Given:

\begin{aligned} &\cos ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\} \\ &-1

Solution:

$y=\cos ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\} \\$

Let,

$x=\sin \theta$

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

\begin{aligned} &\mathrm{y}=\cos ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\} \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{\sin \theta+\sqrt{\cos ^{2} \theta}}{\sqrt{2}}\right\} \\ &\text { Using } \sin ^{2} \theta+\cos ^{2} \theta=1 \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\} \\ &\mathrm{y}=\cos ^{-1}\left\{\sin \theta \frac{1}{\sqrt{2}}+\cos \theta \frac{1}{\sqrt{2}}\right\} \\ &\mathrm{y}=\cos ^{-1}\left\{\sin \theta \sin \left(\frac{\mathrm{\pi}}{4}\right)++\cos \theta \cos \left(\frac{\mathrm{\pi}}{4}\right)\right\} \end{aligned}

$\therefore \sin \left ( \frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$

$\cos \left ( \frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$

Using,

\begin{aligned} &\cos (A-B)=\cos A \cos B+\sin A \sin B \\ &y=\cos ^{-1}\left\{\cos \left(\theta-\frac{\pi}{4}\right)\right\} \end{aligned}

Considering the limits,

\begin{aligned} &-1

Now,

\begin{aligned} &y=\cos ^{-1}\left\{\cos \left(\theta-\frac{\pi}{4}\right)\right\} \\ &y=-\left(\theta-\frac{\pi}{4}\right) \end{aligned}                                                $\left\{\therefore \cos ^{-1}(\cos \theta)=-0 \text { if } \theta \varepsilon[-\pi, 0]\right\}$

\begin{aligned} &y=-\left(\theta-\frac{\pi}{4}\right) \\ &y=-\sin ^{-1} x+\frac{\pi}{4} \end{aligned}                                                                                    $\left [ since\: x=\sin \theta \right ]$

Differentiating with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(-\sin ^{-1} x+\frac{\pi}{4}\right) \\ &\therefore \frac{d}{d x}(\text { constants })=0 \\ &\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}}+0 \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}

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