How does one go about differentiating x^cos(x) ?
tried doing it using chain rule. not sure though.
Let $\displaystyle y = x^{cos(x)}$
Then
$\displaystyle ln(y) = ln(x^{cos(x)}) = cos(x)~ln(x)$
Now take the derivative:
$\displaystyle \frac{1}{y} \frac{dy}{dx} = -sin(x)~ln(x) + cos(x) \cdot \frac{1}{x}$
So
$\displaystyle \frac{dy}{dx} = y \left ( \frac{cos(x)}{x} - sin(x)~ln(x) \right )$
and we know what y is so,
$\displaystyle \frac{dy}{dx} = (x^{cos(x)} ) \left ( \frac{cos(x)}{x} - sin(x)~ln(x) \right )$
-Dan