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If $\sin (x+y)+\cos (x+y)=\sin \left[\cos ^{-1}\left(\frac{1}{3}\right)\right]$, then $\frac{d y}{d x}=$
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Let $\cos ^{-1}\left(\frac{1}{3}\right)=\theta \Rightarrow \cos \theta=\frac{1}{3} \Rightarrow \sin \theta=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$
$\therefore \sin (x+y)+\cos (x+y)=\sin \left[\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\right] \Rightarrow \sin (x+y)+\cos (x+y)=\frac{2 \sqrt{2}}{3}$
Differentiating both sides. w.r.t. $\mathrm{x}$
$\begin{array}{l}
\cos (x+y)\left(1+\frac{d y}{d x}\right)-s \\
x+y=\tan ^{-1}(1)=\frac{\pi}{4}
\end{array}$
Differentiating w.r.t. $x$, we get
$1+\frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-1$
Let $\cos ^{-1}\left(\frac{1}{3}\right)=\theta \Rightarrow \cos \theta=\frac{1}{3} \Rightarrow \sin \theta=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$
$\therefore \sin (x+y)+\cos (x+y)=\sin \left[\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\right] \Rightarrow \sin (x+y)+\cos (x+y)=\frac{2 \sqrt{2}}{3}$
Differentiating both sides. w.r.t. $\mathrm{x}$
$\begin{array}{l}
\cos (x+y)\left(1+\frac{d y}{d x}\right)-s \\
x+y=\tan ^{-1}(1)=\frac{\pi}{4}
\end{array}$
Differentiating w.r.t. $x$, we get
$1+\frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-1$
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