Question

differentiate x^sinx w.r.t x​

Answer 1

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Answer 2

$y = {x}^{sinx} \\ \\ applying \: logarithm \\ \\ logy = sinxlogx \\ \\ \frac{1}{y} \frac{dy}{dx} = sinx \frac{1}{x} + logxcosx \\ \\ \frac{dy}{dx} = y( \frac{sinx}{x} + logxcosx) \\ \\ \frac{dy}{dx} = {x}^{sinx} ( \frac{sinx}{x} + logxcosx)$