Show that $d/dx (a^x) = a^x\ln a$.

Solution 1:

Hint: $a^x=e^{\ln a^x}=e^{x\ln a}$.

Solution 2:

Let $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x)=a^x$ and consider the $\ln$ function. We can take the composition so that we have:

$$(\ln\circ f)(x)=\ln (a^x)=x\ln a$$

Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:

$$\frac{f'(x)}{f(x)}=\ln a$$

Now solving for $f'(x)$ gives $f'(x) = f(x) \ln a$ so that $f'(x) = a^x \ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.

Solution 3:

Hint: Write $y = a^x$ or equivalently $\ln y = x \ln a$ and use implicit differentiation.

Solution 4:

, the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.