Confusion in understanding logarithmic differentiation

Question

Suppose $y=u^v$ then $dy\over dx$$=v*u^{v-1}=$$dy\over dx$$1\over y$$=logu^v$

Does this means that the power which $e$ will be raised so that it is equal to $u^v$ would be equal to $u^v$

Answer 1

I'm not sure that I fully understand what you trying to do. But in order to find the derivative write $$ u^v = e^{v \log u} $$ and then take the derivative.

The other way (equivalent to the first one) is to take the logarithm $$ \log y = v \log u $$ and then differentiate $$ \frac{y^{'}}{y} = (v \log u)^{'} \implies y^{'} = y \cdot (v \log u)^{'}. $$

Answer 2

$$y=u^v\iff \log y=v\log u$$ so that

$$(\log y)'=\frac{y'}y=v'\log u+v\frac{u'}{u}$$

and

$$y'=\left(v'\log u+v\frac{u'}{u}\right)u^v.$$