# How do you implicitly differentiate 2=e^(xy)-ln(x^3y)+siny ?

##### 1 Answer
Feb 26, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 y - x {y}^{2} {e}^{x y}}{{x}^{2} y {e}^{x y} - x + x y \cos y}$

#### Explanation:

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$. However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

$2 = {e}^{x y} - \ln \left({x}^{3} y\right) + \sin y$

First let's simplify the expression using the rules for logarithms:

$\setminus \setminus \setminus \setminus \setminus {e}^{x y} - \ln \left({x}^{3}\right) - \ln \left(y\right) + \sin y = 2$
$\therefore {e}^{x y} - 3 \ln \left(x\right) - \ln \left(y\right) + \sin y = 2$

Differentiate wrt $x$:

${e}^{x y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \frac{3}{x} - \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$y {e}^{x y} + x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{3}{x} - \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$x {y}^{2} {e}^{x y} + {x}^{2} y {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y - x \frac{\mathrm{dy}}{\mathrm{dx}} + x y \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

${x}^{2} y {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} + x y \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 3 y - x {y}^{2} {e}^{x y}$

$\left({x}^{2} y {e}^{x y} - x + x y \cos y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 3 y - x {y}^{2} {e}^{x y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 y - x {y}^{2} {e}^{x y}}{{x}^{2} y {e}^{x y} - x + x y \cos y}$

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

So Let $F \left(x , y\right) = {e}^{x y} - \ln \left({x}^{3} y\right) + \sin y - 2$; Then:

$\frac{\partial F}{\partial x} = y {e}^{x y} - \frac{3}{x} \setminus \setminus$; and $\setminus \setminus \frac{\partial F}{\partial x} = x {e}^{x y} - \frac{1}{y} + \cos y$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y {e}^{x y} - \frac{3}{x}}{x {e}^{x y} - \frac{1}{y} + \cos y}$

$\setminus \setminus \setminus \setminus \setminus = - \frac{\frac{1}{x} \left(x y {e}^{x y} - 3\right)}{\frac{1}{y} \left(x y {e}^{x y} - 1 + y \cos y\right)}$

$\setminus \setminus \setminus \setminus \setminus = - \frac{y}{x} \frac{\left(x y {e}^{x y} - 3\right)}{\left(x y {e}^{x y} - 1 + y \cos y\right)}$

$\setminus \setminus \setminus \setminus \setminus = \frac{3 y - x {y}^{2} {e}^{x y}}{\left({x}^{2} y {e}^{x y} - x + x y \cos y\right)}$

Here we get an immediate implicit function for the derivative, so here if you are familiar with partial differentiation the solution is much easier