If the given function is complicated to solve in order to obtain an inverse, you can use:
\(\displaystyle \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\)
\(\displaystyle x=(2y-\pi)^{3}+2y-cos(y)\)
\(\displaystyle \frac{dx}{dy}=6(2y-\pi)^{2}+2+sin(y)\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{6(2y-\pi)^{2}+2+sin(y)}\)
Leave it as is. It is too complicated to solve for y in terms of x.
Or, implicit differentiation can be used:
\(\displaystyle \frac{d}{dx}[x]=\frac{d}{dx}[(2y-\pi)^{3}+2y-cos(y)]\)
\(\displaystyle 1=(6(2y-\pi)^{2}\frac{dy}{dx}+2+sin(y))\frac{dy}{dx}\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{6(2y-\pi)^{2}+2+sin(y)}\)
Which agrees with the first case.
