# How do you use implicit differentiation to find (dy)/(dx) given sin2x^2y^3=3x^3+1?

Jul 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{9 {x}^{2} - 4 x \cos \left(2 {x}^{2}\right) {y}^{3}}{3 \sin \left(2 {x}^{2}\right) {y}^{2}}$

#### Explanation:

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

However, we only differentiate explicit functions of $y$ wrt $x$. But if we apply the chain rule we can differentiate an implicit function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Example:

$\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dy}} \left({y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Now, we have:

$\sin 2 {x}^{2} {y}^{3} = 3 {x}^{3} + 1$

Implicitly differentiating wrt $x$ (applying product rule):

$\left(\sin 2 {x}^{2}\right) \left(\frac{d}{\mathrm{dx}} {y}^{3}\right) + \left(\frac{d}{\mathrm{dx}} \sin 2 {x}^{2}\right) \left({y}^{3}\right) = 9 {x}^{2}$

$\therefore \left(\sin 2 {x}^{2}\right) \left(3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(4 x \cos 2 {x}^{2}\right) \left({y}^{3}\right) = 9 {x}^{2}$

$\therefore 3 \sin \left(2 {x}^{2}\right) {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x \cos \left(2 {x}^{2}\right) {y}^{3} = 9 {x}^{2}$

$\therefore 3 \sin \left(2 {x}^{2}\right) {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 9 {x}^{2} - 4 x \cos \left(2 {x}^{2}\right) {y}^{3}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{9 {x}^{2} - 4 x \cos \left(2 {x}^{2}\right) {y}^{3}}{3 \sin \left(2 {x}^{2}\right) {y}^{2}}$

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

So Let $F \left(x , y\right) = \sin 2 {x}^{2} {y}^{3} - 3 {x}^{3} - 1$; Then;

$\frac{\partial F}{\partial x} = 4 x \cos \left(2 {x}^{2}\right) {y}^{3} - 9 {x}^{2}$

$\frac{\partial F}{\partial y} = \sin \left(2 {x}^{2}\right) \left(3 {y}^{2}\right)$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 x \cos \left(2 {x}^{2}\right) {y}^{3} - 9 {x}^{2}}{\sin \left(2 {x}^{2}\right) \left(3 {y}^{2}\right)}$
$\text{ } = \frac{9 {x}^{2} - 4 x \cos \left(2 {x}^{2}\right) {y}^{3}}{3 \sin \left(2 {x}^{2}\right) {y}^{2}}$, as before.