How do you find the derivative of $x^(sinx)$ ?

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Answer 1 · 2016-11-17T20:31:37

$d/dx x^sinx= x^sinx((sinx)/x -cosxlnx)$

Let $y = x^sinx$

Then $ln y = ln(x^sinx)$

$:. lny = (sinx)lnx$

Differentiating implicitly and applying the product rule:

$1/ydy/dx=(sinx)(1/x) + (-cosx)(lnx)$
$1/ydy/dx=(sinx)/x -cosxlnx$
$dy/dx=y((sinx)/x -cosxlnx)$
$dy/dx=x^sinx((sinx)/x -cosxlnx)$

How do you find the derivative of x^(sinx)x^(sinx)?