Implicit derivative

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

[math]\displaystyle{ \frac{\mathrm{d}t}{\mathrm{d}x} = \frac{\mathrm{d}t}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x} }[/math]

If we let [math]\displaystyle{ t = f(y) }[/math], then,

[math]\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} }[/math]

Example

[math]\displaystyle{ x = 6y^2 + 5x^4 - y^3 }[/math]

[math]\displaystyle{ 1 = 6\frac{\mathrm{d}}{\mathrm{d}x}y^2 + 20x^3 - \frac{\mathrm{d}}{\mathrm{d}x}y^3 }[/math]

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

[math]\displaystyle{ 1 - 20x^3 = 6\cdot2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3y^2\frac{\mathrm{d}y}{\mathrm{d}x} }[/math]

Then we can isolate [math]\displaystyle{ \frac{\mathrm{d}y}{\mathrm{d}x} }[/math]

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

Then divide to get,

[math]\displaystyle{ \frac{1 - 20x^3}{12y - 3y^2} = \frac{\mathrm{d}y}{\mathrm{d}x} }[/math]