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If y=e2log⁡ety=e^{2\log_e t} and x=log⁡eet2x=\log_e e^{t^2}, then the value of d2ydx2\frac{d^2y}{dx^2} is

Solution

✅ Correct Option: 1

y=e2log⁡et=t2y=e^{2\log_e t}=t^2 and x=log⁡eet2=t2x=\log_e e^{t^2}=t^2.

So y=xy=x, giving dydx=1\frac{dy}{dx}=1 and d2ydx2=0\frac{d^2y}{dx^2}=0.

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