# Question c760c

Sep 14, 2017

$\mathrm{dx} - \left(y \mathrm{dx} + x \mathrm{dy}\right) \setminus \frac{1}{{\sin}^{2} \left(x y\right)} = 0$

#### Explanation:

Explicit differentiation is when you have y = f(x), that is y and x are separated.
Implicit differentiation is needed when you have y = f(x,y). The differentiation will connect dy to x and to dy. You will have to process that equation further to fully separate dy and x.

The strategy is to differentiate, that is find the differential, not the derivative.

Differentiate the first term, $d \left(x\right) = \mathrm{dx}$.

Adding the second term, we first differentiate with respect to $x y$ then with respect to x and y. Because y appears on both sides of the equation, the differentiation is implicit.

$\mathrm{dx} - d \cot x y = \left[\mathrm{dx} - \setminus \frac{\setminus \partial \left\{x y\right\}}{\setminus \partial \left(x\right)} \mathrm{dx} + \setminus \frac{\setminus \partial x y}{\setminus \partial y} \mathrm{dy}\right] d \left(\cot \frac{x y}{d} \left(x y\right)\right] = \mathrm{dx} - \left(y \mathrm{dx} + x \mathrm{dy}\right) \setminus \frac{1}{{\sin}^{2} \left(x y\right)} = 0$

Sep 15, 2017

DIvide by dx. you get
1 - (y + xdy/dx)1/sin^2(xy) ] 0#

Then isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ to obtain

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

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