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Logarithmic Differentiation

Differentiate...

Question

Differentiate
$log (x^{x} - 2^{sin x})$

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Solution

$y = l o g (x^{x} - 2^{s i n x})$
$e^{y} = x^{x} - 2^{s i n x}$
Diffferentiating both sides wrt x
$e^{y} \frac{d y}{d x} = x^{x} (1 + l o g x) - 2^{s i n x} c o s x . l o g 2$
$\frac{d y}{d x} = \frac{x^{x} (1 + l o g x) - 2^{s i n x} c o s x . l o g 2}{x^{x} - 2^{s i n x}}$

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