If $x=e^{y+e^{y+ ........ to \,\, &infin

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

If $x=e^{y+e^{y+ ........ to \,\, ∞}}, x> 0$ then $\frac{d^2y}{dx^2}$ is :

Options:

$-x^2$

$-\frac{1}{x^2}$

$\frac{1}{x^2}$

$x^2$

Correct Answer:

$-\frac{1}{x^2}$

Explanation:

The correct answer is Option (2) → $-\frac{1}{x^2}$

$x=e^{y+e^{y+ ........ to \,\, ∞}}$

$x=e^{y+x}⇒\log x=y+x$

so $y=\log x=x$

$\frac{dy}{dx}=\frac{1}{x}-1⇒\frac{d^2y}{dx^2}=-\frac{1}{x^2}$