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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Aicharya · Accepted Answer

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

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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