Thread: Derivative of a general function

1. Derivative of a general function

Hello,

I have to find the derivative of a general function $\displaystyle f(x,y)$ where $\displaystyle x(s,t) = \frac{s+t}{2}$ and $\displaystyle y(s,t) = \frac{s-t}{2}$.

I have already determined that:
$\displaystyle \frac{\partial f}{\partial s} = \frac{1}{2}(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})$
and
$\displaystyle \frac{\partial f}{\partial t} = \frac{1}{2}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y})$

So, my question is, how can I use this information to get a value for $\displaystyle \frac{\partial d^2f}{\partial s \partial t}$, again for a general function?

Thanks for your help in advance, if any clarification is need please ask.

2. Originally Posted by jonmondalson
Hello,

I have to find the derivative of a general function $\displaystyle f(x,y)$ where $\displaystyle x(s,t) = \frac{s+t}{2}$ and $\displaystyle y(s,t) = \frac{s-t}{2}$.

I have already determined that:
$\displaystyle \frac{\partial f}{\partial s} = \frac{1}{2}(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})$
and
$\displaystyle \frac{\partial f}{\partial t} = \frac{1}{2}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y})$

So, my question is, how can I use this information to get a value for $\displaystyle \frac{\partial^2f}{\partial s \partial t}$, again for a general function?
To find the second derivatives with respect to s or t, use the chain rule, just as you did to find the first derivatives. In fact,

\displaystyle \begin{aligned}\frac{\partial^2f}{\partial s \partial t} = \frac{\partial}{\partial s}\Bigl(\frac{\partial f}{\partial t}\Bigr) &= \frac{\partial}{\partial x}\Bigl(\frac{\partial f}{\partial t}\Bigr)\frac{\partial x}{\partial s} + \frac{\partial}{\partial y}\Bigl(\frac{\partial f}{\partial t}\Bigr)\frac{\partial y}{\partial s} \\ &= \frac{\partial}{\partial x}\Bigl(\tfrac{1}{2}\bigl(\tfrac{\partial f}{\partial x} - \tfrac{\partial f}{\partial y}\bigr)\Bigr)*\tfrac12 + \frac{\partial}{\partial y}\Bigl(\tfrac{1}{2}\bigl(\tfrac{\partial f}{\partial x} - \tfrac{\partial f}{\partial y}\bigr)\Bigr)*\tfrac12 \\ &= \ldots \text{ \footnotesize (I'll leave you to finish it)} .\end{aligned}