Finding Maclaurin’s series expansion function of $f(x) = a^x$ at $x=0$

Question

First derivation is easy as it's $a^x * \log a$, but I have some troubles with finding the derivative of $ a^x * \log a$

Using the product rule I have $ (a^x *\log a)' = (a^x * \log^2 a) + a^x * \frac{1}{a}$, whereas wolfram says it should be just $a^x * \log^2 a$.

Need a hint.

Answer 1

Hint: Differentiation is with respect to $x $.

Answer 2

We get $$(a^x)'=a^x\log(a)$$ $$(a^x\log(a))'=a^x(\log(a))^2$$ $$(a^x(\log(a))^2)'=a^x(\log(a))^3$$ etc.

Answer 3

Note that $$ a^x=e^{x\log{a}} $$ You know how to take derivatives of the exponential.