# Partial Derivative/Examples/u + ln u = x y

## Example of Partial Derivative

Let $u + \ln u = x y$ be an implicit function.

Then:

 $\displaystyle \dfrac {\partial u} {\partial x}$ $=$ $\displaystyle \dfrac {u y} {u + 1}$ $\displaystyle \dfrac {\partial u} {\partial y}$ $=$ $\displaystyle \dfrac {u x} {u + 1}$

## Proof

 $\displaystyle u + \ln u$ $=$ $\displaystyle x y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac \partial {\partial x} u + \dfrac \partial {\partial x} \ln u$ $=$ $\displaystyle \dfrac \partial {\partial x} x y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\partial u} {\partial x} + \dfrac 1 u \dfrac {\partial u} {\partial x}$ $=$ $\displaystyle y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\partial u} {\partial x} \paren {1 + \dfrac 1 u}$ $=$ $\displaystyle y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\partial u} {\partial x}$ $=$ $\displaystyle \dfrac {u y} {u + 1}$

and:

 $\displaystyle u + \ln u$ $=$ $\displaystyle x y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac \partial {\partial y} u + \dfrac \partial {\partial y} \ln u$ $=$ $\displaystyle \dfrac \partial {\partial y} x y$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\partial u} {\partial y} + \dfrac 1 u \dfrac {\partial u} {\partial y}$ $=$ $\displaystyle x$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\partial u} {\partial y} \paren {1 + \dfrac 1 u}$ $=$ $\displaystyle x$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\partial u} {\partial y}$ $=$ $\displaystyle \dfrac {u x} {u + 1}$

$\blacksquare$