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

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

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

Example

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

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

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

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

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

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

Then divide to get,

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

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