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

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

Answers (1)

Answer:

\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)

 

Posted by

infoexpert26

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