How do you calculate y^2x = 2sin(xy^3) For implicit differentiation?

May 17, 2018

The answer is $= \frac{y \left(2 y \cos \left(x {y}^{3} - 1\right)\right)}{2 x \left(1 - 3 y \sin \left(x {y}^{3}\right)\right)}$

Explanation:

Let

$f \left(x , y\right) = {y}^{2} x - 2 \sin \left(x {y}^{3}\right)$

Then,

$\frac{\partial f}{\partial x} = {y}^{2} - 2 {y}^{3} \cos \left(x {y}^{3}\right) = {y}^{2} \left(1 - 2 y \cos \left(x {y}^{3}\right)\right)$

$\frac{\partial f}{\partial y} = 2 y x - 6 x {y}^{2} \sin \left(x {y}^{3}\right) = 2 y x \left(1 - 3 y \sin \left(x {y}^{3}\right)\right)$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = - \frac{{y}^{2} \left(1 - 2 y \cos \left(x {y}^{3}\right)\right)}{2 y x \left(1 - 3 y \sin \left(x {y}^{3}\right)\right)}$

$= \frac{y \left(2 y \cos \left(x {y}^{3}\right) - 1\right)}{2 x \left(1 - 3 y \sin \left(x {y}^{3}\right)\right)}$