# How do you implicitly differentiate x/y^2-x=1/sin(x)+y?

##### 1 Answer

$y ' = \frac{{y}^{3} \cdot \csc x \cdot \cot x + y - {y}^{3}}{{y}^{3} + 2 x}$

#### Explanation:

The given equation is

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

differentiate both sides of the equation with respect to x

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

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

$\frac{{y}^{2} \cdot 1 - x \cdot 2 y \cdot y '}{{y}^{4}} - 1 = \frac{\sin x \cdot \left(0\right) - 1 \cdot \left(\cos x\right)}{{\sin}^{2} x} + y '$

$\frac{{y}^{2} - 2 x \cdot y \cdot y '}{{y}^{4}} - 1 = \frac{- \cos x}{{\sin}^{2} x} + y '$

Take note: $\frac{- \cos x}{{\sin}^{2} x} = - \csc x \cdot \cot x$

$\frac{{y}^{2}}{y} ^ 4 - \frac{2 x \cdot y \cdot y '}{{y}^{4}} - 1 = - \csc x \cdot \cot x + y '$

$\frac{1}{y} ^ 2 - \frac{2 x y '}{{y}^{3}} - 1 = - \csc x \cdot \cot x + y '$

Using transposition, it follows

$- y ' - \frac{2 x y '}{{y}^{3}} = 1 - \csc x \cdot \cot x - \frac{1}{y} ^ 2$

Multiply both sides of the equation by ${y}^{3}$

${y}^{3} \left[- y ' - \frac{2 x y '}{{y}^{3}}\right] = {y}^{3} \left[1 - \csc x \cdot \cot x - \frac{1}{y} ^ 2\right]$

$- {y}^{3} y ' - 2 x y ' = {y}^{3} - {y}^{3} \cdot \csc x \cdot \cot x - y$

Factor the $y '$ and dividing both sides by $\left(- {y}^{3} - 2 x\right)$

$\frac{\left(- {y}^{3} - 2 x\right) y '}{\left(- {y}^{3} - 2 x\right)} = \frac{{y}^{3} - {y}^{3} \cdot \csc x \cdot \cot x - y}{- {y}^{3} - 2 x}$

$\frac{\cancel{\left(- {y}^{3} - 2 x\right) y '}}{\cancel{\left(- {y}^{3} - 2 x\right)}} = \frac{{y}^{3} - {y}^{3} \cdot \csc x \cdot \cot x - y}{- {y}^{3} - 2 x}$

$y ' = \frac{{y}^{3} - {y}^{3} \cdot \csc x \cdot \cot x - y}{- {y}^{3} - 2 x}$

$y ' = \frac{{y}^{3} \cdot \csc x \cdot \cot x + y - {y}^{3}}{{y}^{3} + 2 x}$

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