# How do you differentiate x-cos(x^2)+y^2/x+3x^5=4x^3?

Oct 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} + 12 {x}^{4} - 15 {x}^{6} - {x}^{2} - 2 {x}^{3} \sin \left({x}^{2}\right)}{2 x y}$

#### Explanation:

We have:

$x - \cos \left({x}^{2}\right) + {y}^{2} / x + 3 {x}^{5} = 4 {x}^{3}$

Method 1: Implicit differentiation, as is:

Differentiating wrt $x$ whilst applying the chain rule and the product rule:

$1 + \sin \left({x}^{2}\right) \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) + \frac{\left(x\right) \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) - \left({y}^{2}\right) \left(\frac{d}{\mathrm{dx}} x\right)}{x} ^ 2 + 15 {x}^{4} = 12 {x}^{2}$

$\therefore 1 + \sin \left({x}^{2}\right) \left(2 x\right) + \frac{\left(x\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left({y}^{2}\right) \left(1\right)}{x} ^ 2 + 15 {x}^{4} = 12 {x}^{2}$

$\therefore 1 + 2 x \sin \left({x}^{2}\right) + \frac{2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2}}{x} ^ 2 + 15 {x}^{4} = 12 {x}^{2}$

$\therefore \frac{2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2}}{x} ^ 2 = 12 {x}^{2} - 15 {x}^{4} - 1 - 2 x \sin \left({x}^{2}\right)$

$\therefore 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2} = 12 {x}^{4} - 15 {x}^{6} - 1 - 2 {x}^{3} \sin \left({x}^{2}\right)$

$\therefore 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} + 12 {x}^{4} - 15 {x}^{6} - {x}^{2} - 2 {x}^{3} \sin \left({x}^{2}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} + 12 {x}^{4} - 15 {x}^{6} - {x}^{2} - 2 {x}^{3} \sin \left({x}^{2}\right)}{2 x y}$

Method 2: Implicit Function Theorem

Putting:

$F \left(x , y\right) = x - \cos \left({x}^{2}\right) + {y}^{2} / x + 3 {x}^{5} - 4 {x}^{3}$

We have:

$\frac{\partial F}{\partial x} = 1 + 2 x \sin \left({x}^{2}\right) - {y}^{2} / {x}^{2} + 15 {x}^{4} - 12 {x}^{2}$

$\frac{\partial F}{\partial y} = \frac{2 y}{x}$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{1 + 2 x \sin \left({x}^{2}\right) - {y}^{2} / {x}^{2} + 15 {x}^{4} - 12 {x}^{2}}{\frac{2 y}{x}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{\left(x\right) \left(1 + 2 x \sin \left({x}^{2}\right) - {y}^{2} / {x}^{2} + 15 {x}^{4} - 12 {x}^{2}\right)}{2 y}$

Yielding the same result as above.