If y=x+e^{x}, then {d² x}{d y²} is equal to

Correct Answer is : −ex(1+ex)3{-e^{x}}{≤ft(1+e^{x})³}(1+ex)3−ex

Correct Answer is : −ex(1+ex)3{-e^{x}}{≤ft(1+e^{x})³}(1+ex)3−ex

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MHTCET 13 Apr 2026 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

y=x+e^{x}

\frac{d^{2} x}{d y^{2}}

is equal to

Solution:

Concept:

To find

\frac{d^{2} x}{d y^{2}}

, use the inverse derivative

\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}

and apply the chain rule.

Explanation:

Given

y = x + e^{x}

Differentiate with respect to

x

\frac{dy}{dx} = 1 + e^{x}

Therefore,

\frac{dx}{dy} = \frac{1}{1 + e^{x}}

Now,

\frac{d^{2}x}{dy^{2}} = \frac{d}{dy}\left( \frac{dx}{dy} \right)

Use

\frac{d}{dy} = \frac{d}{dx} \cdot \frac{dx}{dy}

\frac{d^{2}x}{dy^{2}} = \frac{d}{dx}\left( \frac{1}{1 + e^{x}} \right) \cdot \frac{dx}{dy}

Differentiate

\frac{1}{1+e^{x}}

with respect to

x

\frac{d}{dx}\left( \frac{1}{1+e^{x}} \right) = \frac{-e^{x}}{\left(1+e^{x}\right)^{2}}

Multiply by

\frac{dx}{dy} = \frac{1}{1+e^{x}}

\frac{d^{2}x}{dy^{2}} = \frac{-e^{x}}{\left(1+e^{x}\right)^{2}} \cdot \frac{1}{1+e^{x}}

\frac{d^{2}x}{dy^{2}} = \frac{-e^{x}}{\left(1+e^{x}\right)^{3}}

Answer:

The correct option is D:

\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}

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