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If \(y=f(x)\) is twice differentiable function such that at a point \(P, \frac{d y}{d x}=4, \frac{d^2 y}{d x^2}=-3\), then \(\left(\frac{d^2 x}{d y^2}\right)_P=\)
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Verified Answer
The correct answer is:
\(\frac{3}{64}\)
\(\begin{array}{ll}
\because \frac{d^2 x}{d y^2} =\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d x}{d y} \frac{d}{d x}\left(\frac{1}{\left(\frac{d y}{d x}\right)}\right) \\
=\left(\frac{d x}{d y}\right)\left(\frac{-\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^2}\right)=-\frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d y}{d x}\right)^3} \\
\therefore\left(\frac{d^2 x}{d y^2}\right)_p =-\frac{(-3)}{(4)^3}=\frac{3}{64}
\end{array}\)
Hence, option (4) is correct.
\because \frac{d^2 x}{d y^2} =\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d x}{d y} \frac{d}{d x}\left(\frac{1}{\left(\frac{d y}{d x}\right)}\right) \\
=\left(\frac{d x}{d y}\right)\left(\frac{-\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^2}\right)=-\frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d y}{d x}\right)^3} \\
\therefore\left(\frac{d^2 x}{d y^2}\right)_p =-\frac{(-3)}{(4)^3}=\frac{3}{64}
\end{array}\)
Hence, option (4) is correct.
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