Partial Derivates Question How would I go about taking the partial derivative of:
$\frac{a+b^2}{a^2b}$ with respect to $a$ or $b$?
Thanks!
 A: The partial derivative of $\frac{a+b^2}{a^2b}$ with respect to $a$ is computed by treating $b$ as a constant. 
Thus, $\dfrac{\partial u}{\partial a}[\frac{a+b^2}{a^2b}]$ = $\frac{a^2b(1) - (a+b^2)(2ab)}{a^4b^2}$ = $\frac{a^2b - 2a^2b - 2ab^3}{a^4b^2}$ = $\frac{-a-2b^2}{a^3b}$.
The partial derivative of $\frac{a+b^2}{a^2b}$ with respect to $b$ is computed by treating $a$ as a constant. 
Thus, $\dfrac{\partial u}{\partial b}[\frac{a+b^2}{a^2b}]$ = $\frac{a^2b(2b) - (a+b^2)(a^2)}{a^4b^2}$ = $\frac{2a^2b^2-a^3-a^2b^2}{a^4b^2}$ = $\frac{a^2b^2-a^3}{a^4b^2}$=$\frac{b^2-a}{a^2b^2}$.
A: You have a function of two variables $f(a,b)$. If you want take the partial derivative with respect to $a$, you should assume that the other variables are constants in the function (in this case the $b$ variable). So you can see this like take other function $g(a)=f(a,b)$ with $b$ be a constant in $g$ and take derivative of $g$ in "standard way". To derive with respect to $b$, you can do the same with $a$ be a constant.
