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If t=e2xt=e^{2 x} and y=log⁡et2y=\log _{e} t^{2}, then d2ydx2\frac{d^{2} y}{d x^{2}} is :

Solution

✅ Correct Option: 1

To find d2ydx2\dfrac{d^2y}{dx^2} when t=e2xt=e^{2x} and y=log⁡et2y=\log_e t^2:

First, let's substitute the value of tt into the expression for yy:

y=log⁡et2=log⁡e(e2x)2=log⁡ee4x=4xy = \log_e t^2 = \log_e (e^{2x})^2 = \log_e e^{4x} = 4x

This uses the property log⁡eea=a\log_e e^a = a.


Finding the first derivative:

Since y=4xy = 4x

dydx=4\dfrac{dy}{dx} = 4


Finding the second derivative:

Since dydx=4\dfrac{dy}{dx} = 4 (a constant)

d2ydx2=0\dfrac{d^2y}{dx^2} = 0


Alternatively, we could verify using the chain rule:

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

dydt=2t\dfrac{dy}{dt} = \dfrac{2}{t}

dtdx=2e2x=2t\dfrac{dt}{dx} = 2e^{2x} = 2t

dydx=dydt⋅dtdx=2t⋅2t=4\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{2}{t} \cdot 2t = 4

This confirms our answer that d2ydx2=0\dfrac{d^2y}{dx^2} = 0

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