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If $x^{2} y^{5}=(x+y)^{7}$, then $\frac{d^{2} y}{d x^{2}}$ is equal to
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Given, $x^{2} y^{5}=(x+y)^{7}$
Taking log on both sides, we get
$$
2 \log x+5 \log y=7 \log (x+y)
$$
On differentiating, we get
$$
\begin{aligned}
\frac{2}{x}+\frac{5}{y} \frac{d y}{d x} &=\frac{7}{x+y}\left(1+\frac{d y}{d x}\right) \\
\Rightarrow \quad & \frac{d y}{d x}\left(\frac{7}{x+y}-\frac{5}{y}\right)=\frac{2}{x}-\frac{7}{x+y}
\end{aligned}
$$
$$
\Rightarrow \quad \frac{d y}{d x}=\frac{y}{x}
$$
Again, differentiating, we get
$$
\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=\frac{x \frac{d y}{d x}-y}{x^{2}} \\
&=\frac{x \cdot(y / x)-y}{x^{2}}[\text { from Eq. (i)] }\\
&=0
\end{aligned}
$$
Taking log on both sides, we get
$$
2 \log x+5 \log y=7 \log (x+y)
$$
On differentiating, we get
$$
\begin{aligned}
\frac{2}{x}+\frac{5}{y} \frac{d y}{d x} &=\frac{7}{x+y}\left(1+\frac{d y}{d x}\right) \\
\Rightarrow \quad & \frac{d y}{d x}\left(\frac{7}{x+y}-\frac{5}{y}\right)=\frac{2}{x}-\frac{7}{x+y}
\end{aligned}
$$
$$
\Rightarrow \quad \frac{d y}{d x}=\frac{y}{x}
$$
Again, differentiating, we get
$$
\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=\frac{x \frac{d y}{d x}-y}{x^{2}} \\
&=\frac{x \cdot(y / x)-y}{x^{2}}[\text { from Eq. (i)] }\\
&=0
\end{aligned}
$$
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