First slide

Derivatives

Question

$If y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x \dots \dots \propto}}} then \frac{d y}{d x} is equal to$

Moderate

A

$\frac{x}{2 y - 1}$
B

$\frac{x}{2 y + 1}$
C

$\frac{1}{x (2 y - 1)}$
D

$\frac{1}{x (1 - 2 y)}$

Solution

$\begin{array}{l} y = \sqrt{\log x + y} \Rightarrow y^{2} = \log x + y \Rightarrow y^{2} - y = \log x \\ (2 y - 1) \frac{d y}{d x} = \frac{1}{x} \\ \frac{d y}{d x} = \frac{1}{x (2 y - 1)} \end{array}$

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