# Find dy/dx for the function e^(cos z) + e^(sin y) =1/4 ?

Feb 22, 2018

$\text{see explanation}$

#### Explanation:

$\text{differentiate "color(blue)"implicitly with respect to x}$

$\Rightarrow {e}^{\cos x} \left(- \sin x\right) + {e}^{\sin y} \left(\cos y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\Rightarrow - \sin x {e}^{\cos x} + \cos y {e}^{\sin y} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos y {e}^{\sin y}\right) = \sin x {e}^{\cos x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin x {e}^{\cos x}}{\cos y {e}^{\sin y}}$

Feb 23, 2018

We can apply the Implicit Function Theorem:

If we define:

$f \left(x , y\right) = {e}^{\cos z} + {e}^{\sin y} - \frac{1}{4}$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / {f}_{y}$

And we calculate the partial derivatives:

${f}_{x} = \frac{\partial f}{\partial x}$
$\setminus \setminus \setminus = {e}^{\cos x} \left(- \sin x\right)$

${f}_{y} = \frac{\partial f}{\partial y}$
$\setminus \setminus \setminus = {e}^{\sin y} \left(\cos y\right)$

So then:

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