If x · _{e}≤ft( _{e} x)-x²+y²=4(y > 0), then d y/d x at x=e is

Correct Answer is : 2e−124+e2 {2 e-1}{2 {4+e²}}24+e22e−1

Correct Answer is : 2e−124+e2 {2 e-1}{2 {4+e²}}24+e22e−1

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MHTCET 5 May 2025 Shift 2 Paper

Question : 131 of 150

Marks: +1, -0

x \cdot \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y > 0)

\frac{d y}{d x}

x=e

Solution:

Concept:

Implicit differentiation is used to find

\frac{dy}{dx}

from the given equation, followed by substitution of

x=e

and the corresponding value of

y

Explanation:

Given:

x \log_e(\log_e x) - x^2 + y^2 = 4

with

y > 0

Differentiate both sides with respect to

x

\log_e(\log_e x) + x \cdot \frac{1}{\log_e x} \cdot \frac{1}{x} - 2x + 2y \frac{dy}{dx} = 0

Simplify:

\log_e(\log_e x) + \frac{1}{\log_e x} - 2x + 2y \frac{dy}{dx} = 0

Thus,

\frac{dy}{dx} = \frac{2x - \log_e(\log_e x) - \frac{1}{\log_e x}}{2y}

Now, at

x = e

\log_e e = 1

and

\log_e(\log_e e) = \log_e 1 = 0

Therefore,

\left(\frac{dy}{dx}\right)_{x=e} = \frac{2e - 0 - 1}{2y} = \frac{2e - 1}{2y}

Find

y

x=e

from the original equation:

e \cdot 0 - e^2 + y^2 = 4 \Rightarrow y^2 = 4 + e^2

Since

y > 0

y = \sqrt{4 + e^2}

Substitute this value of

y

\left(\frac{dy}{dx}\right)_{x=e} = \frac{2e - 1}{2\sqrt{4 + e^2}}

Answer:

\frac{dy}{dx}\biggr|_{x=e} = \frac{2e - 1}{2\sqrt{4 + e^2}}

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