# How do you differentiate cosx^2/y^2-xy=y/x^2?

Mar 1, 2016

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

#### Explanation:

$\cos {x}^{2} / {y}^{2} - x y = \frac{y}{x} ^ 2$

$y = \left(\cos {x}^{2} / {y}^{2} - x y\right) {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos {x}^{2} / {y}^{2} - x y\right) {x}^{2}$

Applying product rule as: ${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = \setminus \frac{\setminus \cos \left({x}^{2}\right)}{{y}^{2}} - x y , g = {x}^{2}$

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

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

(Applying sum/difference rule as: ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$
$= \setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{\setminus \cos \left({x}^{2}\right)}{{y}^{2}}\right) - \setminus \frac{d}{\mathrm{dx}} \left(x y\right)$
Here,
$= \setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{\setminus \cos \left({x}^{2}\right)}{{y}^{2}}\right)$ = $- \setminus \frac{2 x \setminus \sin \left({x}^{2}\right)}{{y}^{2}}$
and,
$\setminus \frac{d}{\mathrm{dx}} \left(x y\right)$ = $y$)

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

Again,
$\setminus \frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 x$

Finally,
$= \left(- \setminus \frac{2 x \setminus \sin \left({x}^{2}\right)}{{y}^{2}} - y\right) {x}^{2} + 2 x \left(\setminus \frac{\setminus \cos \left({x}^{2}\right)}{{y}^{2}} - x y\right)$

Simplifying it,

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