Question

show that if u and v are differentiable functions of x and u> ​​0, then the y = u ^ ---> y '= u ^ v (v' * ln (v * u '/ u) ?

Answer 1

was that supposed to be y = u ^ v `implies` other stuff?

Answer 2

\[\large y=u^v~~\iff~~\ln y=v\ln u\]
Differentiate both sides with respect to \(x\):
\[\frac{d}{dx}\ln y=\frac{d}{dx}v\ln u\\
\frac{1}{y}y'=\frac{d}{dx}[v]\ln u+v\frac{d}{dx}[\ln u]\\
\frac{1}{y}y'=v'\ln u+v\frac{1}{u}u'\\
~~~~~~~~~~~~\vdots\]