How do you differentiate $y = x^{cos x}$ $y=x^cosx$ ?

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Answer 1 · 2015-06-06T23:26:07

$y = x^{cos (x)}$ $y=x^cos(x)$

Take logarithm of each side.
$ln (y) = ln (x^{cos (x)})$ $ln(y)=ln(x^cos(x))$

$ln (y) = cos (x) ln (x)$ $ln(y)=cos(x)ln(x)$

Take derivative of each side

$\frac{d}{d x} [ln (y)] = \frac{d}{d x} [cos (x) ln (x)]$ $d/dx[ln(y)]=d/dx[cos(x)ln(x)]$

$\frac{1}{y} \frac{d y}{d x} = - sin (x) ln (x) + \frac{cos (x)}{x}$ $1/ydy/dx=-sin(x)ln(x)+cos(x)/x$

$\frac{d y}{d x} = y (- sin (x) ln (x) + \frac{cos (x)}{x})$ $dy/dx=y(-sin(x)ln(x)+cos(x)/x )$

Substitute $y$ $y$

$\frac{d y}{d x} = x^{cos (x)} (- sin (x) ln (x) + \frac{cos (x)}{x})$ $dy/dx=x^cos(x)(-sin(x)ln(x)+cos(x)/x )$

How do you differentiate y=x^cosxy=xcosxy=x^cosx?