# How do you implicitly differentiate 2=e^(xy)cosxy ?

Jun 18, 2018

$y ' = - \frac{y}{x}$

#### Explanation:

By the chain rule we get

$0 = {e}^{x y} \left(y + x y '\right) \cos \left(x y\right) + {e}^{x y} \left(- \sin \left(x y\right)\right) \left(y + x y '\right)$
$0 = {e}^{x y} y \cos \left(x y\right) + {e}^{x y} x y ' \cos \left(x y\right) - {e}^{x y} y \sin \left(x y\right) - {e}^{x y} x y ' \sin \left(x y\right)$
$0 = {e}^{x y} y \left(\sin \left(x y\right) - \cos \left(x y\right)\right) = {e}^{x y} x \left(\cos \left(x y\right) - \sin \left(x y\right)\right) y '$

dividing by ${e}^{x y} \ne 0$

then we get

$- y \left(\cos \left(x y\right) - \sin \left(x y\right)\right) = x \left(\cos \left(x y\right) - \sin \left(x y\right)\right) y '$
if $\sin \left(x y\right) - \cos \left(x y\right) \ne 0$
we get

$y ' = - \frac{y}{x}$ and $x \ne 0$

Jun 18, 2018

$y ' = - \frac{y}{x}$

#### Explanation:

Let's write $y$ as $y \left(x\right)$, to remind that it is a function of $x$.

Let $u \left(x\right) = x y \left(x\right)$.

$\therefore {e}^{u \left(x\right)} \cos \left(u \left(x\right)\right) = 2$

Differentiate both sides with respect to $x$ and use the product rule:

[e^(u(x))]'cos(u(x))+e^(u(x))[cos((u(x))]'=0

Using the chain rule, we get that:

$\left[{e}^{u \left(x\right)}\right] ' = u ' \left(x\right) {e}^{u \left(x\right)}$

$\left[\cos \left(u \left(x\right)\right)\right] ' = - \sin \left(u \left(x\right)\right) \cdot u ' \left(x\right)$

As such,

$u ' \left(x\right) {e}^{u \left(x\right)} \cos \left(u \left(x\right)\right) - u ' \left(x\right) {e}^{u \left(x\right)} \sin \left(u \left(x\right)\right) = 0$

u'(x)e^(u(x))(cos(u(x))-sin(u(x))=0

Finally, the derivative of $u$ is

$u ' \left(x\right) = \left[x y \left(x\right)\right] ' = x ' y \left(x\right) + x y ' \left(x\right) = y \left(x\right) + x y ' \left(x\right)$

As such, the derivative of our function is

$\left(y \left(x\right) + x y ' \left(x\right)\right) {e}^{x y \left(x\right)} \left(\cos \left(x y \left(x\right)\right) - \sin \left(x y \left(x\right)\right)\right) = 0$

$\left(y + x \left(\text{d"y)/("d} x\right)\right) {e}^{x y} \left(\cos x y - \sin x y\right) = 0$

$y + x \left(\text{d"y)/("d"x)=0 => ("d"y)/("d} x\right) = - \frac{y}{x}$