How to prove that $\dfrac{d}{dx}e^{kx}=ke^{kx}$ So how to prove that  $\dfrac{d}{dx}e^{kx}=ke^{kx}$?
What I did is to think about $e^{kx}$ not as a single function but as a composition of functions, that is $h(x)=e^{kx}=f[g(x)]=f[e^x]=e^{kx}$ where $f(x)=x^k$ and $g(x)=e^x.$ So now I can use the chain rule! aka the Outside-Inside rule! and I get that $h'(x)=f'[g(x)]\cdot g'(x)$. But then I struggle in finding the derivative of $f[g(x)]$ with respect to $g(x)$, if somebody could ever help me, I will be so thankful!
Thank you!
 A: Here, let's put $f(x) = e^x$ and $g(x) = kx$. Then the derivative is equal to $$f'(g(x))\cdot g'(x) = e^{g(x)}g'(x) =  e^{kx}\cdot k = ke^{kx}$$
by the chain rule.
In general, the derivative of $e^{g(x)}$ is equal to $$e^{g(x)}\cdot g'(x)$$
In this case, $g(x) = kx$
A: $\frac{d}{dx}(e^x)^k=k(e^x)^{k-1}(e^x)'=k(e^{kx-x}).e^x=ke^{kx-x+x}=ke^{kx}$
A: Let $y=e^x$, then $h(x)=y^k$
You then have $\cfrac {dh(x)}{dy}=ky^{k-1}$ and $\cfrac {dy}{dx}=y=e^x$ so that $$\frac{dh}{dx}=\frac{dh}{dy}\cdot\frac{dy}{dx}=ky^{k-1}\cdot y=ky^k=ke^{kx}$$
The alternative method from amWhy would be the normal way to do it.
A: $\frac{d}{dx}e^{kx}=\frac{d}{d(kx)}e^{kx} \frac{d}{dx}{kx}$
A: Assuming the chain rule and the fact that $\dfrac{d}{dx}(e^x)=e^x$, this is almost immediate.
We can write $e^{kx}=f(g(x))$, where $f(u)=e^u$ and $g(x) = kx$.
Now $f'(g(x))$ is what you get when you calculate $f'(u)$ and substitute $u=g(x)$.
Can you see where to take it from here?
A: I believe it would be easier for you to switch the functions you are differentiating: $ e^{kx} = f(g(x)) $ where $ f(x) = e^x, ~ g(x) = kx $. By chain rule, $ [f(g(x))]' = f'(g(x))\cdot g'(x) = e^{kx}\cdot k $
