# Implicit derivative

derivative of functions

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

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

If we let ${\displaystyle t=f(y)}$, then,

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

## Example

${\displaystyle x=6y^{2}+5x^{4}-y^{3}}$
${\displaystyle 1=6{\frac {\mathrm {d} }{\mathrm {d} x}}y^{2}+20x^{3}-{\frac {\mathrm {d} }{\mathrm {d} x}}y^{3}}$

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

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

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

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

Then divide to get,

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