Question 39439
We don't have a commonly used theorem for this kind of operation (such as the product rule, etc.).
We can get the g(x) out of the exponent if we take the ln of the function, but we have to undo the damage by taking the exponential (i.e. the inverse of ln) of the result.  Note that  exp(ln x)=x is normally written like this:
{{{e^(ln(x))=x}}}
Following this pattern we can write:
{{{f(x)^g(x)}}} = {{{e^(ln(f(x)^g(x)))}}} = {{{e^(g(x)*ln(f(x)))}}}
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Now we can differentiate using the chain rule and product rule:
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{{{d(e^(g(x)*ln(f(x))))/dx}}} = {{{(e^(g(x)*ln(f(x))))*((g(x)/(f(x)))*(d(f(x))/(dx))+ln(f(x))*(d(g(x))/(dx)))))}}}