First slide

Derivatives

Question

If $x = e^{y + e^{y + \dots \infty}}$ , $x > 0, then \frac{dy}{dx}$ is equal to

Easy

A

$\frac{x}{1 + x}$
B

$\frac{1}{x}$
C

$\frac{1 - x}{x}$
D

$\frac{1 + x}{x}$

Solution

given that, $x = e^{y + e^{y + \dots \cdot n}}$
$∴ x = e^{y + x}$
Taking log on both sides, we get
$\log x = (y + x)$
On differentiating w.r.t. x, we get
$\begin{aligned} \frac{1}{x} = \frac{dy}{dx} + 1 \\ \Rightarrow \frac{dy}{dx} = \frac{1 - x}{x} \end{aligned}$

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