CUET UG 2024 — Mathematics Calculus

To find $\dfrac{d^2y}{dx^2}$dx2d2y​ when $t=e^{2x}$t=e2x and $y=\log_e t^2$y=loge​t2:

First, let's substitute the value of $t$t into the expression for $y$y:
$$y = \log_e t^2 = \log_e (e^{2x})^2 = \log_e e^{4x} = 4x$$y=loge​t2=loge​(e2x)2=loge​e4x=4x

This uses the property $\log_e e^a = a$loge​ea=a.

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Finding the first derivative:

Since $y = 4x$y=4x

$$\dfrac{dy}{dx} = 4$$dxdy​=4

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Finding the second derivative:

Since $\dfrac{dy}{dx} = 4$dxdy​=4 (a constant)

$$\dfrac{d^2y}{dx^2} = 0$$dx2d2y​=0

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Alternatively, we could verify using the chain rule:
$$y = \log_e t^2 = 2\log_e t$$y=loge​t2=2loge​t

$$\dfrac{dy}{dt} = \dfrac{2}{t}$$dtdy​=t2​

$$\dfrac{dt}{dx} = 2e^{2x} = 2t$$dxdt​=2e2x=2t

$$\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{2}{t} \cdot 2t = 4$$dxdy​=dtdy​⋅dxdt​=t2​⋅2t=4

This confirms our answer that $\dfrac{d^2y}{dx^2} = 0$dx2d2y​=0