# How do you find (dy)/(dx) given xe^siny=e^y?

Oct 29, 2016

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

#### Explanation:

The implicit differentiation of the given equation is determined by applying some differentiation properties

color(blue)((u+v)'=(du)/dxxxv+(dv)/dxxxu

$\textcolor{red}{\frac{{\mathrm{de}}^{u}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \times {e}^{u}}$

$\textcolor{b r o w n}{\frac{\mathrm{ds} \in y}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \times \cos y}$

Start computing $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

$\Rightarrow \textcolor{b l u e}{\frac{\mathrm{dx}}{\mathrm{dx}} \times \left({e}^{\sin y}\right) + \left(\frac{{\mathrm{de}}^{\sin} y}{\mathrm{dx}}\right) \times x} = \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} \times {e}^{y}}$

$\Rightarrow {e}^{\sin} y + \textcolor{red}{\frac{\mathrm{ds} \in y}{\mathrm{dx}}} \times {e}^{\sin} y \times x = \frac{\mathrm{dy}}{\mathrm{dx}} \times {e}^{y}$

$\Rightarrow {e}^{\sin} y + \textcolor{red}{\frac{\mathrm{ds} \in y}{\mathrm{dx}}} \times {e}^{\sin} y \times x = \frac{\mathrm{dy}}{\mathrm{dx}} \times {e}^{y}$

$\Rightarrow {e}^{\sin} y + \textcolor{b r o w n}{\frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos y\right)} {e}^{\sin} y x = \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y}$

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

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

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

Oct 29, 2016

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

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

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

$\ln x + \sin y = y$

Use use normal and implicit differentiation...

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

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

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

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

Oct 29, 2016

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

#### Explanation:

I would solve the problem by taking natural logarithms;

$x {e}^{\sin} y = {e}^{y}$
$\therefore \ln \left(x {e}^{\sin} y\right) = \ln \left({e}^{y}\right)$
$\therefore \ln x + \ln \left({e}^{\sin} y\right) = \ln {e}^{y}$
$\therefore \ln x + \sin y = y$

Differentiate wrt $x$
$\therefore \frac{d}{\mathrm{dx}} \ln x + \frac{d}{\mathrm{dx}} \sin y = \frac{d}{\mathrm{dx}} y$

$\therefore \frac{1}{x} + \frac{d}{\mathrm{dy}} \sin y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

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

Oct 29, 2016

$\frac{1}{x \left(1 - \cos y\right)}$

#### Explanation:

This can be organized as

$x = {e}^{y} / \left(e \left(\sin y\right)\right) = {e}^{y - \sin y}$. So,

$y ' = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}}$

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

=1/(x(1-cos y),