Question

f(x)=√(e^2x+1) then f'(0)=?

Answer 1

What do you get when you apply the chain rule twice to this?

Answer 2

well first I rewrote f(x) to be \[f(x)=((e^2x)+1)\]^(1/2), then I tried the chain rule but i forgot how to find the derivative of e^2x

Answer 3

\[(e^{f(x)})'=f'(x)e^{f(x)}\]

Answer 4

Okay, so after you applied chain rule once, you got \[f'(x)=\frac{1}{2}(e^{2x}+1)^{-0.5}\frac{d}{dx}(e^{2x}+1)\] Correct? (Just wanted to make sure we're on the same page) If the derivative of \[e^u\] is itself, what does a second chain rule tell you?

Answer 5

So the derivative of e^2x is e^2x? I completlely understand the chain rulue though

Answer 6

I'm trying to find the second part to the equation the d/dx(e^2x+1)

Answer 7

Why just e^(2x)? There's more to it than that. You must use chain rule! Set u=2x and go from there. Once you solve this part, you're basically done.

Answer 8

I understand the rest. I was just confused with that portion.

Answer 9

Okay, if you want to find \[\frac{d}{dx}(e^{2x}+1)=\frac{d}{dx}e^{2x}+\frac{d}{dx}1=\frac{d}{dx}e^{2x}\] This is just an application of the chain rule. Since \[\frac{d}{dx}e^{2x}=e^{2x}\frac{d}{dx}(2x)=2e^{2x}\]

Answer 10

okay that's it! :)