$\require{cancel}$
I've started watch MIT Courseware Single Variable Calculus; there Is a lesson on how to find out how to find the derivative of a function with the power as x: $f(x) = a^x $.
The lecturer starts of by show that $\frac{d}{dx} a^x = \lim_{x \to 1} \ \frac{a^{x+\Delta x} – a^x}{\Delta x} = a^x\frac{(a^{\Delta x} -1)}{\Delta x} $
$a^x \lim_{x \to 1} \left[\frac{(a^{\Delta x} -1)}{\Delta x}\right] \to M(a)$
so: $$\frac{d}{dx} \ a^x = a^x M(a)$$ and we don't know what $M(a)$ is yet.
but the derivative at $x = 0$ of $a^x $ is going to be $M(a)$ (that is the slope at x = 0 is going to be $M(a)$).
he then propopes that there is a base $e$ where $M(e) = 1$ so that the derivative of $ \frac{d}{dx} \ e ^x = e^x$.
and the idea is that he is going to show why $e$ exists
$$f(x) = 2^x $$ stretch function by $k$
$$f(kx) = 2^{kx} = {2^k}^x = b^x$$
$$b = 2^k$$
$$\frac{d}{dx}b^x = \frac{d}{dx}f(kx) = kf ^\prime(kx)$$ I don't quite understand how $\frac{d}{dx}f(kx) = kf ^\prime(kx)$. He uses the chain rule but I don't quite see how that works. isn't $f^\prime(x)$ the same as $\frac{d}{dx}f(x)$
He also adds that the same thing is done when taking the derivative of $sin(kt)$
$$\frac{d}{dt}sin(kt) = kcos(kt)$$. I understand the $sin()$ example as the chain rule because $kt$ is a composition of $sin()$ so we can do
$$\frac{d}{dx}sin(kt) = \frac{d}{\cancel{dx}}sin(x).\frac{\cancel{dx}}{dt}kt$$
how does the chain rule apply to something with a power
Best Answer
$$\frac{d}{dx}f(kx) = kf ^\prime(kx)$$
is an application of the following rule: $$[f(g(x))]'=f'(g(x))\times g'(x)$$