L'Hopitals Rule with derivatives?

[m0286]

Question is:
Evaluate the following limits
lim x--> 0 4^x - e^2x / 2x
So i take derivatives
but that's where I am confused... Whats the derivate of 4^x... x4^-x?
and is the derivative of -e^2x -2e^x?
So then that leaves me with 4x^-x -2e^x /2which is 0... so do i do derivatives again? It seems like the x's won't go away?

 

[StatusX]

You can use the chain rule and the fact that the derivative of e^x is e^x to answer both questions. That is, the derivative of e^f(x) is e^f(x)*f'(x). Try to rewrite 4^x as e^f(x). (use logs)

 

[Jameson]

$$f(x)=4^x$$
$$\ln(f(x))=x\ln(4)$$
$$\frac{1}{f(x)}*f'(x)=\ln(4)$$

You can take it from there.