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If $x=t^2$ and $y=t^3$, then $\frac{d^2 y}{d x^2}$ is equal to
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Verified Answer
The correct answer is:
$\frac{3}{4 t}$
$\frac{3}{4 t}$
Since, $x=t^2$ and $y=t^3$
So, $\frac{d x}{d t}=2 t$ and $\frac{d y}{d t}=3 t^2$
...(i)
and $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3 t^2}{2 t}=\frac{3}{2} t$
Again differentiating w.r.t. $x$, we have
$$
\begin{aligned}
&\frac{d^2 y}{d x^2}=\frac{3}{2}\left(\frac{d}{d t} t\right)\left(\frac{d t}{d x}\right)=\frac{3}{2}\left(\frac{1}{2 t}\right) \quad \text { [From eq. (i)] } \\
&=\frac{3}{4 t}
\end{aligned}
$$
So, $\frac{d x}{d t}=2 t$ and $\frac{d y}{d t}=3 t^2$
...(i)
and $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3 t^2}{2 t}=\frac{3}{2} t$
Again differentiating w.r.t. $x$, we have
$$
\begin{aligned}
&\frac{d^2 y}{d x^2}=\frac{3}{2}\left(\frac{d}{d t} t\right)\left(\frac{d t}{d x}\right)=\frac{3}{2}\left(\frac{1}{2 t}\right) \quad \text { [From eq. (i)] } \\
&=\frac{3}{4 t}
\end{aligned}
$$
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