# How do you use implicit differentiation to find dy/dx given e^(xsiny)=y?

Oct 7, 2016

${e}^{x \sin y} = y$

Rewrite as:

${e}^{\square} = y$

Where $\square = x \sin y$.

Let's find the derivative of $\square$.

$\square ' = 1 \times \sin y + x \times \cos y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\square ' = \sin y + x \cos y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

The derivative of $y = {e}^{f} \left(x\right)$ is always $f ' \left(x\right) \times {e}^{f \left(x\right)}$, so $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sin y + x \cos y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) {e}^{x \sin y}$

However, to determine the complete derivative of the function, we need to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation.

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - x \cos y {e}^{x \sin y}\right) = \sin y {e}^{x \sin y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin y {e}^{x \sin y}}{1 - x \cos y {e}^{x \sin y}}$

Substitute the initial function as $y$ for ${e}^{x \sin y}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \sin y}{1 - x y \cos y}$

Hopefully this helps!

Oct 7, 2016

#### Explanation:

${e}^{x \sin y} = y$

$\ln y = x \sin y$

Now use both implicit differentiation formulas:

1. On the left... $\frac{\mathrm{dy}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

2. On the right... $f \left(x\right) g ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + g \left(y\right) f ' \left(x\right)$

If you do this, what you'll get is...

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \sin y$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - x \cdot \cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sin y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left\{\frac{1}{y} - x \cdot \cos y\right\} = \sin y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left\{\frac{1}{y} - \frac{x y \cos y}{y}\right\} = \sin y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left\{\frac{1 - x y \cos y}{y}\right\} = \sin y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin y \cdot \frac{y}{1 - x y \cos y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \sin y}{1 - x y \cos y}$