JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

Let

x = {\log _e}t,t > 0

and

y + 1 = {t^2}.

Then,

\frac{{{d^2}x}}{{d{y^2}}}

is equal to

View answer and explanation

Correct option: B

- \frac{1}{2}{e^{ - 4x}}

Option B (

- \frac{1}{2}{e^{ - 4x}}

) is correct because it is the result of differentiating the given equations twice and simplifying.

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x

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[x]

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