# How do you use implicit differentiation to find (dy)/(dx) given x^2=(4x^2y^3+1)^2?

Sep 8, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{12 x {y}^{2} \left(4 {x}^{2} {y}^{3} + 1\right)} - \frac{4 y}{3 x}$

#### Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$.

However, some functions y are written implicitly as functions of $x$. So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

In the given instance, we have ${x}^{2} = {\left(4 {x}^{2} {y}^{3} + 1\right)}^{2}$. Now differentiating both sides w.r.t. $x$ and remembering to multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$, whenever we differentiate w.r.t. $y$, we get

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

or $2 x = \left(16 x {y}^{3} + 12 {x}^{2} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right)$ or

$\left(16 x {y}^{3} + 12 {x}^{2} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{2 x}{2 \left(4 {x}^{2} {y}^{3} + 1\right)} = \frac{x}{4 {x}^{2} {y}^{3} + 1}$ or

$12 {x}^{2} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{4 {x}^{2} {y}^{3} + 1} - 16 x {y}^{3}$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{12 {x}^{2} {y}^{2} \left(4 {x}^{2} {y}^{3} + 1\right)} - \frac{16 x {y}^{3}}{12 {x}^{2} {y}^{2}}$

= $\frac{1}{12 x {y}^{2} \left(4 {x}^{2} {y}^{3} + 1\right)} - \frac{4 y}{3 x}$