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If $x=\log t$ and $y=t^{2}-1$, then what is $\frac{d^{2} y}{d x^{2}}$ at $t=1$ equal to?
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The correct answer is:
4
Let $x=\log t$ and $y=t^{2}-1$
$x=\log t$
$\Rightarrow 2 x=2 \log t$
$\Rightarrow 2 x=\log t^{2}$
$\Rightarrow 2 x=\log (y+1) \Rightarrow e^{2 x}=y+1$
On differentiating w.r.t. $x$, twice, we get
$e^{2 x} 2=\frac{d y}{d x} \Rightarrow 4 e^{2 x}=\frac{d^{2} y}{d x^{2}}$
At $t=1, x=0$
$\frac{d^{2} y}{d x^{2}}=4 e^{2(0)}=4 \quad\left(\because e^{0}=1\right)$
$x=\log t$
$\Rightarrow 2 x=2 \log t$
$\Rightarrow 2 x=\log t^{2}$
$\Rightarrow 2 x=\log (y+1) \Rightarrow e^{2 x}=y+1$
On differentiating w.r.t. $x$, twice, we get
$e^{2 x} 2=\frac{d y}{d x} \Rightarrow 4 e^{2 x}=\frac{d^{2} y}{d x^{2}}$
At $t=1, x=0$
$\frac{d^{2} y}{d x^{2}}=4 e^{2(0)}=4 \quad\left(\because e^{0}=1\right)$
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