Suppose that you want to differentiate a function of the form

$$y(x)=\bigl(f(x)\bigr)^{g(x)}$$ with $f(x)>0$. This can be accomplished by rewriting the function as $$y(x)=e^{g(x)\cdot \ln\bigl(f(x)\bigr)}$$ and applying the chain rule and product rule for differentiation.
Note that you get the result of rewriting of the function by applying the natural logarithm. This gives $\ln\bigl(y(x)\bigr)=\ln\left( \bigl(f(x)\bigr)^{g(x)}\right)=g(x)\cdot \ln\bigl(f(x)\bigr)$. Apply the exponential function on both sides of the equation and simplify the left-hand side.

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But finding the derivative $y'(x)$ can also be done via the method known under the name logarithmic differentiation. Here you first apply the natural logarithm on either side of the expression; you will then

$$\ln\bigl(y(x)\bigr)=g(x)\cdot \ln\bigl(f(x)\bigr)\tiny.$$ Hereafter you compute the derivatives of the expressions on the left- and right-hand side of the equal sign; this leads via the rules for differentiation to $$\frac{y'(x)}{y(x)}=g'(x)\cdot \ln\bigl(f(x)\bigr)+ g'(x)\cdot \frac{f'(x)}{f(x)}\tiny.$$ So $$\begin{aligned}y'(x)&=y(x)\cdot\left(g'(x)\cdot \ln\bigl(f(x)\bigr)+ g'(x)\cdot \frac{f'(x)}{f(x)}\right)\\ \\ &= \bigl(f(x)\bigr)^{g(x)}\cdot \left(g'(x)\cdot \ln\bigl(f(x)\bigr)+ g'(x)\cdot \frac{f'(x)}{f(x)}\right)\tiny.\end{aligned}$$

More instructive is the following example of logarithmic differentiation.