# How do you implicitly differentiate -3=(x^2+y)^3-y^2x ?

Apr 28, 2017

$\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x {\left({x}^{2} + y\right)}^{2} - {y}^{2}}{2 x y - 3 {\left({x}^{2} + y\right)}^{2}}}$

#### Explanation:

Implicit differentiation is basically done in cases where $y$ cannot be explicitly written as a function of $x$.

In this case,

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

Differentiating both sides w.r.t. $x$

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

Using chain rule to evaluate $\frac{d {\left({x}^{2} + y\right)}^{3}}{\mathrm{dx}}$ & product rule to evaluate $\frac{d \left({y}^{2} x\right)}{\mathrm{dx}}$

$\implies 0 = \frac{d {\left({x}^{2} + y\right)}^{3}}{d \left({x}^{2} + y\right)} \cdot \frac{d \left({x}^{2} + y\right)}{\mathrm{dx}} - \left[{y}^{2} \frac{\mathrm{dx}}{\mathrm{dx}} + x {\mathrm{dy}}^{2} / \mathrm{dx}\right]$

$\implies {y}^{2} \frac{\mathrm{dx}}{\mathrm{dx}} + x {\mathrm{dy}}^{2} / \mathrm{dx} = \frac{d {\left({x}^{2} + y\right)}^{3}}{d \left({x}^{2} + y\right)} \cdot \frac{d \left({x}^{2} + y\right)}{\mathrm{dx}}$

Using sum rule to evaluate $\frac{d \left({x}^{2} + y\right)}{\mathrm{dx}}$ & chain rule to evaluate ${\mathrm{dy}}^{2} / \mathrm{dx}$

$\implies {y}^{2} + x \cdot {\mathrm{dy}}^{2} / \mathrm{dy} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \left[3 \cdot {\left({x}^{2} + y\right)}^{2}\right] \cdot \left[{\mathrm{dx}}^{2} / \mathrm{dx} + \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

$\implies {y}^{2} + x \cdot 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\left({x}^{2} + y\right)}^{2} \cdot \left[2 x + \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

$\implies {y}^{2} + 2 x y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x {\left({x}^{2} + y\right)}^{2} + 3 {\left({x}^{2} + y\right)}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies 2 x y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {\left({x}^{2} + y\right)}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x {\left({x}^{2} + y\right)}^{2} - {y}^{2}$

$\implies \left[2 x y - 3 {\left({x}^{2} + y\right)}^{2}\right] \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x {\left({x}^{2} + y\right)}^{2} - {y}^{2}$

$\implies \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x {\left({x}^{2} + y\right)}^{2} - {y}^{2}}{2 x y - 3 {\left({x}^{2} + y\right)}^{2}}}$