Statement
The derivative of a nested function is the derivative of the outer function multiplied by the derivative of the inner function.
$$ \tfrac{d}{dx}\bigg(g(j(x))\bigg) = g'(j(x)) * j'(x) $$
Proof
Define function $ f $.
$$ f(x) = g(j(x)) $$
Take the definition for the derivative.
$$ f'(x) = \lim_{h \to 0}\left(\frac{f(x + h) - f(x)}{h}\right) $$
$$ f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{h}\right) $$
Multiply the numerator and denominator of the fraction by $ j(x + h) - j(x) $ and split the limit.
$$ f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{h} * \frac{j(x + h) - j(x)}{j(x + h) - j(x)}\right) $$
$$ f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{j(x + h) - j(x)} * \frac{j(x + h) - j(x)}{h}\right) $$
$$ f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{j(x + h) - j(x)}\right) * \lim_{h \to 0}\left(\frac{j(x + h) - j(x)}{h}\right) $$
Let $ k = j(x + h) - j(x) $. From this follows that $ j(x + h) = k + j(x) $.
Now we can conclude that as $ h $ approaches to $ 0 $, so does $ k $:
$$ k = j(x + h) - j(x) = j(x + 0) - j(x) = j(x) - j(x) = 0 $$
Substitute $ k $ in the derivative.
$$ f'(x) = \lim_{k \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{k}\right) * \lim_{h \to 0}\left(\frac{j(x + h) - j(x)}{h}\right) $$
Here are the definitions of the derivative for $ g(j(x)) $ and $ j(x) $.
$$ f'(x) = g'(j(x)) * j'(x) $$