Question

Differentiate the following w.r.tx
$y = \frac{log \: x}{x + log \: x}$
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Answer 1

Step-by-step explanation:

We have,

$\tt{y = \dfrac{ log(x) }{x + log(x) } }$

$\tt{ \implies \: xy + y \: log(x) = log(x)}$

Differentiating both sides w.r.t x,

$\tt{ \implies \: y + x \dfrac{dy}{dx} + y \cdot \dfrac{1}{x} + log(x) \dfrac{dy}{dx} = \dfrac{1}{x} }$

$\tt{ \implies \: y + \dfrac{y}{x} +(x + log(x) ) \dfrac{dy}{dx} = \dfrac{1}{x} }$

$\tt{ \implies \: (x + log(x) ) \dfrac{dy}{dx} = \dfrac{1}{x} - y -\dfrac{y}{x} }$

$\tt{ \implies \: (x + log(x) ) \dfrac{dy}{dx} = \dfrac{1 - xy - y}{x} }$

$\tt{ \implies \: \dfrac{dy}{dx} = \dfrac{1 - xy - y}{x(x + log(x))} }$