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#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 29 maths textbook solution

$\frac{d y}{d x}=\frac{\pi}{4}(\sqrt{2}+1)$

Hint:

Use chain rule and product rule

Given:

$\sin ^{2} y+\cos x y=K$

Solution:

$\sin ^{2} y+\cos x y=K$

Differentiate the given equation w.r.t x

$\frac{d}{d x}\left(\sin ^{2} y+\cos x y\right)=\frac{d(k)}{d x}$

$\frac{d \sin ^{2} y}{d x}+\frac{d \cos x y}{d x}=0$                        $\left[\because \frac{d(\operatorname{con} s \tan t)}{d x}=0\right]$

$\left(\frac{d \sin ^{2} y}{d \sin y} \times \frac{d \sin y}{d y} \times \frac{d y}{d x}\right)+\left(\frac{d \cos x y}{d x y}\right) \times \frac{d x y}{d x}=0$             [Using chain rule]

$2 \sin y \times \cos y \times \frac{d y}{d x}+(-\sin x y) \times\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]=0$            [Using product rule]

$\sin 2 y \frac{d y}{d x}-\sin x y\left(x \frac{d y}{d x}\right)-y \sin x y=0$

$\frac{d y}{d x}(\sin 2 y-x \sin x y)=y \sin x y$

$\frac{d y}{d x}=\frac{y \sin x y}{\sin 2 y-x \sin x y}$

At $x=1, y=\frac{\pi}{4}$

$\frac{d y}{d x}=\frac{\left(\frac{\pi}{4}\right) \sin \left(1 \times \frac{\pi}{4}\right)}{\sin \left(2 \times \frac{\pi}{4}\right)-1 \times \sin \left(1 \times \frac{\pi}{4}\right)}$

$=\frac{\frac{\pi}{4} \cdot\left(\frac{1}{\sqrt{2}}\right)}{\sin \frac{\pi}{2}-\frac{1}{\sqrt{2}}}$                            $\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin \frac{\pi}{2}=1\right]$

$=\frac{\frac{\pi}{4 \sqrt{2}}}{1-\frac{1}{\sqrt{2}}}$

$=\frac{\frac{\pi}{4 \sqrt{2}}}{\frac{(\sqrt{2}-1)}{\sqrt{2}}}$

$=\frac{\pi \times \sqrt{2}}{(\sqrt{2}-1) 4 \sqrt{2}}=\frac{\pi}{4(\sqrt{2}-1)} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\pi(\sqrt{2}+1)}{4}$

Hence at $x=1, y=\frac{\pi}{4}$

$\frac{d y}{d x}=\frac{\pi}{4}(\sqrt{2}+1)$