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If $x=f(t)$ and $y=g(t)$, then the value of $\frac{d^{2} y}{d x^{2}}$ is
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Verified Answer
The correct answer is:
$\frac{f^{\prime}(t) g^{\prime \prime}(t) g^{\prime}(t) f^{\prime \prime}(t)}{\left\{f^{\prime}(t)\right\}^{3}}$
Given, $x=f(t), y=g(t)$
$$
\begin{array}{c}
\frac{d x}{d t}=f^{\prime}(t), \frac{d y}{d t}=g^{\prime}(t) \\
\therefore \quad \frac{d y}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)}
\end{array}
$$
Now, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$
$$
\begin{aligned}
&=\frac{d}{d t}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \frac{d t}{d x} \\
=&\left[\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^{2}}\right] \cdot \frac{1}{f^{\prime}(t)} \\
=& \frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left\{f^{\prime}(t)\right\}^{3}}
\end{aligned}
$$
$$
\begin{array}{c}
\frac{d x}{d t}=f^{\prime}(t), \frac{d y}{d t}=g^{\prime}(t) \\
\therefore \quad \frac{d y}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)}
\end{array}
$$
Now, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$
$$
\begin{aligned}
&=\frac{d}{d t}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \frac{d t}{d x} \\
=&\left[\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^{2}}\right] \cdot \frac{1}{f^{\prime}(t)} \\
=& \frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left\{f^{\prime}(t)\right\}^{3}}
\end{aligned}
$$
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