# How do you find dy/dx by implicit differentiation of x^2-y^3=0 and evaluate at point (1,1)?

Feb 22, 2017

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{\left(0 , 0\right)} = \frac{2}{3}$

#### Explanation:

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

${x}^{2} - {y}^{3} = 0$

Differentiate wrt $x$:

$2 x - 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \setminus \setminus \setminus \setminus$ .... [1]

Although (in this case) we could find an implicit expression for $\frac{\mathrm{dy}}{\mathrm{dx}}$ it is not always necessary, as we are asked to find the value at a particular point.

At $\left(1 , 1\right)$ (noting that the point does indeed lie on the initial curve) we have (substituting into [1]):

$2 - 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3}$