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If t=e2xt = e^{2x} and log⁡et2\log_e t^2, then d2ydx2\frac{d^2y}{dx^2} is :

Solution

✅ Correct Option: 1

Interpreting y=log⁡et2=2log⁡ety = \log_e t^2 = 2\log_e t.

With t=e2xt = e^{2x}: y=2log⁡e(e2x)=2(2x)=4xy = 2 \log_e(e^{2x}) = 2(2x) = 4x.

dydx=4\frac{dy}{dx} = 4.

d2ydx2=0\frac{d^2y}{dx^2} = 0.

Note: The question text omits 'y =' before log⁡et2\log_e t^2; we have interpreted y=log⁡et2y = \log_e t^2 as the intended relation.

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