First slide

Derivatives

Question

$If \log (x^{2} + y^{2}) = 2 \tan^{- 1} (\frac{y}{x}) then \frac{d y}{d x} =$

Moderate

A

$\frac{x + y}{x - y}$
B

$\frac{x - y}{x + y}$
C

1
D

None

Solution

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

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