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Parametric Differentiation

If x= log t...

Question

If $x =$ log t and $y = t^{2} - 1$ , then $y^{^{''}}$ at $t = 1$ is

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is D $4$
$x = log t$ and $y = t^{2} - 1$
$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 t}{\frac{1}{t}} = 2 t^{2}$

$\frac{d t}{d x} = t$

$\frac{d^{2} y}{d x^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \frac{d t}{d x} = (4 t) t = 4 t^{2}$

$\frac{d^{2} y}{d x^{2}} |_{t = 1} = 4$

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