Me Being Bored

I'm bored so I'm sharing what I believe is a neat proof for something useful. What you need to know to understand this is derivatives, compositions, and a bit of algebra (I think).

Let $g(x)$ be the functional inverse of $f(x)$.

Using knowledge about functional compositions, I can determine that $f(g(x))=x$.

Taking the derivative of both sides...

$\frac{d}{dx}(x)=\frac{d}{dx}(f(g(x)))$

Solving the derivatives...

$1=f'(g(x))g'(x)$

Dividing both sides by $f'(g(x))$...

${f'(g(x))}^{-1}=g'(x)$

${f'(g(x))}^{-1}=g'(x)$ is a simple method that allows an indirect derivation of an inverse function.

Example problem:

$f(x)={2}^{{x}^{2}}$

$f'(x)={2}^{{x}^{2}+1}x*ln(2)$

$g(x)=\sqrt{log_2(x)}$

$\frac{1}{{2}^{{\sqrt{log_2(x)}}^{2}+1}\sqrt{log_2(x)}*ln(2)}=\frac{1}{2*{2}^{log_2(x)}\sqrt{log_2(x)}*ln(2)}=\frac{1}{2x*ln(2)\sqrt{log_2(x)}}=\frac{1}{2x\sqrt{ln(2)*ln(x)}}$

Now checking the actual derivative of the function I get $g'(x)=\frac{1}{2x\sqrt{ln(2)*ln(x)}}$

I'll upload more proofs when I'm bored again.