# How do you find dy/dx by implicit differentiation of x^3+y^3=4xy+1 and evaluate at point (2,1)?

Oct 27, 2017

$\frac{8}{5}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({x}^{3} + {y}^{3} = 4 x y + 1\right) - - \left(1\right)$

we will differentiate $w r t \text{ } x$. Remembering that when differentiating $y$ we multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$ by virtue of the chain rule. Also on the $R H S$ we will need the product rule on the first term

$\left(1\right) \rightarrow 3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 0$

rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 x \frac{\mathrm{dy}}{\mathrm{dx}} = 4 y - 3 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} - 4 x\right) = 4 y - 3 {x}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 y - 3 {x}^{2}}{3 {y}^{2} - 4 x}$

:.[(dy)/(dx)]_(color(white)(=)(2,1))=(4xx1-3xx2^2)/(3xx1^2-4xx2

$= \frac{4 - 12}{3 - 8} = - \frac{8}{-} 5 = \frac{8}{5}$