# How do you find the Implicit differentiation of x^4-5xy^3+y^6=21?

May 17, 2015

You use the fact that $y$ is a function of $x$;

May 18, 2015

Tedious, but worth it:

If $x$ and $y$ are related by:

${x}^{4} - 5 x {y}^{3} + {y}^{6} = 21$,

then we can find $y '$ (or find dy/dx) in either of two ways.

We understand that $y$ is some function (or functions) of $x$. I mean, if we choose an $x$ value, we could find (or at least try to find) the value of $y$ that makes the equation true.

Our choices are 1) make the funcion(s) explicit (obvious), or 2) leave the function(s) implcit (hidden) and find $y '$ without first solving for $y$.

Gio has shown us how to fin $y '$ without first solving for $y$, and that answers the question about implicit differentiation. I want to show the explicit function(s).

We'll solve:
${x}^{4} - 5 x {y}^{3} + {y}^{6} = 21$, for $y$.

${x}^{4} - 5 x {y}^{3} + {y}^{6} = 21$,

${y}^{6} - 5 x {y}^{3} + {x}^{4} - 21 = 0$,

This equation is quadratic in ${y}^{3}$, with $a = 1$, $b = 5 x$, and $c = {x}^{4} - 21$.

Using the quadratic formula, we get:

${y}^{3} = \frac{- \left(- 5 x\right) \pm \sqrt{{\left(- 5 x\right)}^{2} - 4 \left(1\right) \left({x}^{4} - 21\right)}}{2 \left(1\right)}$

${y}^{3} = \frac{5 x \pm \sqrt{25 {x}^{2} - 4 {x}^{4} + 84}}{2}$

So

$y = \sqrt[3]{\frac{5 x \pm \sqrt{25 {x}^{2} - 4 {x}^{4} + 84}}{2}}$

Now, if we wanted to differentiate, we would get the equivalent of

$y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {y}^{3} - 4 {x}^{3}}{6 {y}^{5} - 15 x {y}^{2}}$