+65 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.
