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Integration of Trigonometric Functions

If x=log t ...

Question

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

2

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4

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3

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None of these

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Solution

The correct option is A $4$
$x = log t \Rightarrow \frac{d x}{d t} = \frac{1}{t} y = t^{2} - 1 \Rightarrow \frac{d y}{d t} = 2 t \frac{d y}{d x} = 2 t^{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d}{d t} (2 t^{2}) \times \frac{d t}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 4 t \times t = 4 t^{2}$
$\Rightarrow y^{''} (1) = 4$

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I f x^{2} + y^{2} = t - \frac{1}{t} a n d x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}, t h e n x^{3} y \frac{d y}{d x} e q u a l s

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