# How do you implicitly differentiate -1=y^3x-2xy-9x^4y ?

$y ' = \frac{{y}^{3} - 2 y - 36 {x}^{3} y}{9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}}$

#### Explanation:

From the given equation we differentiate both sides with respect to x

$- 1 = {y}^{3} \cdot x - 2 x y - 9 {x}^{4} \cdot y$

$\frac{d}{\mathrm{dx}} \left(- 1\right) = \frac{d}{\mathrm{dx}} \left({y}^{3} \cdot x - 2 x y - 9 {x}^{4} \cdot y\right)$

$\frac{d}{\mathrm{dx}} \left(- 1\right) = \frac{d}{\mathrm{dx}} \left({y}^{3} \cdot x\right) - 2 \cdot \frac{d}{\mathrm{dx}} \left(x y\right) - 9 \cdot \frac{d}{\mathrm{dx}} \left({x}^{4} \cdot y\right)$

$0 = {y}^{3} \cdot \frac{d}{\mathrm{dx}} \left(x\right) + x \cdot \frac{d}{\mathrm{dx}} \left({y}^{3}\right) - 2 \left[x \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left(x\right)\right] - 9 \left[{x}^{4} \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left({x}^{4}\right)\right]$

$0 = {y}^{3} + x \cdot \left(3 {y}^{2} \cdot y '\right) - 2 \left[x \cdot \left(y '\right) + y \cdot 1\right] - 9 \left[{x}^{4} \left(y '\right) + y \left(4 {x}^{3}\right)\right]$

$0 = {y}^{3} + 3 x \cdot {y}^{2} \cdot y ' - 2 x y ' - 2 y - 9 {x}^{4} y ' - 36 {x}^{3} y$

transposition to obtain

$9 {x}^{4} y ' + 2 x y ' - 3 x \cdot {y}^{2} \cdot y ' = {y}^{3} - 2 y - 36 {x}^{3} y$

factoring to obtain

$\left(9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}\right) y ' = {y}^{3} - 2 y - 36 {x}^{3} y$

divide both sides of the equation by $\left(9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}\right)$

$\frac{\left(9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}\right) y '}{9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}} = \frac{{y}^{3} - 2 y - 36 {x}^{3} y}{9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}}$

$\frac{\cancel{\left(9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}\right)} y '}{\cancel{\left(9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}\right)}} = \frac{{y}^{3} - 2 y - 36 {x}^{3} y}{9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}}$

$y ' = \frac{{y}^{3} - 2 y - 36 {x}^{3} y}{9 {x}^{4} + 2 x - 3 x \cdot {y}^{2}}$

God bless...I hope the explanation is useful.