Implicit derivative

Implicit derivatives are derivatives of implicit functions. This means that they are not in the form of $y=f(x)$ (explicit function), and are instead in the form $0=f(x,y)$ (implicit function). It might not be possible to rearrange the function into the form $y=f(x)$. To use implicit differentiation, we use the chain rule,

${\frac {\mathrm {d} t}{\mathrm {d} x}}={\frac {\mathrm {d} t}{\mathrm {d} y}}{\frac {\mathrm {d} y}{\mathrm {d} x}}$

If we let $t=f(y)$, then,

${\frac {\mathrm {d} }{\mathrm {d} x}}f(y)={\frac {\mathrm {d} }{\mathrm {d} y}}f(y){\frac {\mathrm {d} y}{\mathrm {d} x}}=f'(y){\frac {\mathrm {d} y}{\mathrm {d} x}}$

Example

$x=6y^{2}+5x^{4}-y^{3}$

$1=6{\frac {\mathrm {d} }{\mathrm {d} x}}y^{2}+20x^{3}-{\frac {\mathrm {d} }{\mathrm {d} x}}y^{3}$

Which we can work out to be equivalent to, using the above,

$1-20x^{3}=6\cdot 2y{\frac {\mathrm {d} y}{\mathrm {d} x}}-3y^{2}{\frac {\mathrm {d} y}{\mathrm {d} x}}$

Then we can isolate ${\frac {\mathrm {d} y}{\mathrm {d} x}}$

$1-20x^{3}={\frac {\mathrm {d} y}{\mathrm {d} x}}\left[12y-3y^{2}\right]$

Then divide to get,

${\frac {1-20x^{3}}{12y-3y^{2}}}={\frac {\mathrm {d} y}{\mathrm {d} x}}$