CUET Mathematics 2024 - If t=e^2 x and y= _e t^2, then d^2 yd x^2 is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Continuity & Differentiability

Medium

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

0

4 t

\frac{4 e^{2 t}}{t}

\frac{e^{2 t}(4 t-1)}{t^{2}}

✅ Correct Option: 1

To find

\dfrac{d^2y}{dx^2}

when

t=e^{2x}

and

y=\log_e t^2

First, let's substitute the value of

t

into the expression for

y

\newline

y = \log_e t^2 = \log_e (e^{2x})^2 = \log_e e^{4x} = 4x

This uses the property

\log_e e^a = a

Finding the first derivative:

Since

y = 4x

\dfrac{dy}{dx} = 4

Finding the second derivative:

Since

\dfrac{dy}{dx} = 4

(a constant)

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

Alternatively, we could verify using the chain rule:

\newline

y = \log_e t^2 = 2\log_e t

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

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

\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{2}{t} \cdot 2t = 4

This confirms our answer that

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

Related questions:

CUET Mathematics 2024

f(x)

f(x)=\left\{\begin{array}{lll}k x+1 & \text { if } & x \leq \pi \\ \cos x & \text { if } & x>\pi\end{array}\right.

x=\pi

k

CUET Mathematics 2024

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List-I	List-II
(A) $\|x - 1\| + \|x - 2\|$	(I) is differentiable everywhere except at $x = 0$
(B) $x - \|x\|$	(II) is continuous everywhere
(C) $x - [x]$	(III) is not differentiable at $x = 1$
(D) $x \, \|x\|$	(IV) is differentiable at $x = 1$

CUET Mathematics 2024

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