$y=x^{\cos\ x}$
$\ln\ y = \cos(x)\ln\ x$
$\frac{dy}{dx}\cdot\frac{1}{y}=\frac{-\sin(x)}{x}$
$\frac{dy}{dx}=x^{\cos x}(\frac{-\sin(x)}{x})$
Is my method and steps correct?
$\endgroup$ 42 Answers
$\begingroup$$$y={ x }^{ \cos { x } }\\ \ln { y } =\cos { x } \ln { x } \\ \frac { d\left( \ln { y } \right) }{ dx } =\frac { d\left( \cos { x } \ln { x } \right) }{ dx } \\ \frac { 1 }{ y }\frac { dy }{ dx } =\frac { d\left( \cos { x } \right) }{ dx } \ln { x+\cos { x } \frac { d\left( \ln { x } \right) }{ dx } } \\ \frac { 1 }{ y } \frac { dy }{ dx } =-\sin { x } \ln { x } +\frac { \cos { x } }{ x } \\ \frac { dy }{ dx }={ x }^{ \cos { x } }\left( -\sin { x } \ln { x } +\frac { \cos { x } }{ x } \right) $$
$\endgroup$ 7 $\begingroup$Not quite: $(uv)^\prime = u^\prime v +uv^\prime$, not $u^\prime v^\prime$. The derivative $\frac{d}{dx}(\cos(x)\ln(x))$ is not $\cos^\prime(x)\ln^\prime(x)$.
$\endgroup$ 8
