Suppose you don’t have an explicit expression for a function, but you wish to find an expression for its derivative.  It is sometimes possible to do this using implicit differentiation.  The method is so simple that it hardly deserves a name, but I will present it here because it is a useful method to keep in mind and many calculus textbooks have entire section on the topic.  Suppose you know the function $$!\displaystyle y(x) = f[g(x)]$$ and you need an expression for $$dg/dx$$. One way of proceeding is to invert the equation so that you have $$!\displaystyle g(x)=f^{-1}[y(x)]$$ and then differentiate, but this might me quite difficult in some cases.  It may even be impossible to properly invert the equation. Rather than wasting time with algebra, you can do the following: Differentiate the equation to yield $$!\displaystyle y'(x) = \frac{df}{dg}\frac{dg}{dx}$$ then solve for $$dg/dx$$ $$!\displaystyle \frac{dg}{dx} = \frac{y'(x)}{df/dg}$$ That’s all there is to it!

Example 1: Given $$!\displaystyle 2y^3x+2x^3y-4=0$$ find $$dy/dx$$.

Solution: Differentiate the equation with respect to $$x$$ $$!\displaystyle 2y^3 +6y^2x\frac{dy}{dx} +6x^2y+ 2x^3\frac{dy}{dx}=0$$ now solve for $$dy/dx$$ $$!\displaystyle \frac{dy}{dx} = -\frac{2y^3+6x^2y}{6y^2x+ 2x^3}$$

Example 2: See inverse trig derivatives