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If $\sin (x+y)+\cos (x+y)=\log (x+y)$, then $\frac{d^{2} y}{d x^{2}}=$
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Given that, $\sin (x+y)+\cos (x+y)=\log (x+y)$
On differentiating w.r.t. $x$,
$\begin{aligned}
&\cos (x+y) \cdot\left(1+\frac{d y}{d x}\right)-\sin (x+y)\left(1+\frac{d y}{d x}\right) \\
&=\frac{1}{(x+y)}\left(1+\frac{d y}{d x}\right) \\
&\Rightarrow\left(1+\frac{d y}{d x}\right)\left\{\cos (x+y)-\sin (x+y)-\frac{1}{(x+y)}\right\}=0 \\
&\Rightarrow \quad 1+\frac{d y}{d x}=0
\end{aligned}$
Again differentiating w.r.t. ' $x$ '
$0+\frac{d^{2} y}{d x^{2}}=0 \Rightarrow \frac{d^{2} y}{d x^{2}}=0$
On differentiating w.r.t. $x$,
$\begin{aligned}
&\cos (x+y) \cdot\left(1+\frac{d y}{d x}\right)-\sin (x+y)\left(1+\frac{d y}{d x}\right) \\
&=\frac{1}{(x+y)}\left(1+\frac{d y}{d x}\right) \\
&\Rightarrow\left(1+\frac{d y}{d x}\right)\left\{\cos (x+y)-\sin (x+y)-\frac{1}{(x+y)}\right\}=0 \\
&\Rightarrow \quad 1+\frac{d y}{d x}=0
\end{aligned}$
Again differentiating w.r.t. ' $x$ '
$0+\frac{d^{2} y}{d x^{2}}=0 \Rightarrow \frac{d^{2} y}{d x^{2}}=0$
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