# Partial differentiation

• May 2nd 2011, 02:27 PM
bryan06
Second derivative of partial differentiation???!!
Consider the function of two variables

u(x,y) = f(x-y) + g(x+y/3)

where f(s) and g(t) are each arbitrary functions of a single variable.

Using the change of variables

s=x-y and t=x+y/3

what are the first and second derivatives of u with respect to x and y in terms of f and g??thankssss
(Bow)
• May 2nd 2011, 02:46 PM
TheEmptySet
Quote:

Originally Posted by bryan06
Consider the function of two variables

u(x,y) = f(x-y) + g(x+y/3)

where f(s) and g(t) are each arbitrary functions of a single variable.

Using the change of variables

s=x-y and t=x+y/3

what are the first and second derivatives of u with respect to x and y in terms of f and g??thankssss
(Bow)

Use the chain rule

$\displaystyle \frac{\partial u}{\partial x}=\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial x}$

Now we need to take the 2nd derivative

$\displaystyle \frac{\partial }{\partial x}\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}\right)+\frac{\partial }{\partial x}\left( \frac{\partial f}{\partial t}\frac{\partial t}{\partial x} \right)$

Now we need to use the product rule (and chain) on both of the above terms to get

$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 f}{\partial s^2}\left(\frac{\partial s}{\partial x}\right)^2+\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial x^2}+\frac{\partial^2 f}{\partial t^2}\left(\frac{\partial t}{\partial x}\right)^2+\frac{\partial f}{\partial t}\frac{\partial^2 t}{\partial x^2}$

See if you can finish
• May 2nd 2011, 02:58 PM
bryan06
thanksss~~~yess i did manage to finish it its very kind of you indeed!

Quote:

Originally Posted by TheEmptySet
Use the chain rule

$\displaystyle \frac{\partial u}{\partial x}=\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial x}$

Now we need to take the 2nd derivative

$\displaystyle \frac{\partial }{\partial x}\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}\right)+\frac{\partial }{\partial x}\left( \frac{\partial f}{\partial t}\frac{\partial t}{\partial x} \right)$

Now we need to use the product rule (and chain) on both of the above terms to get

$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 f}{\partial s^2}\left(\frac{\partial s}{\partial x}\right)^2+\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial x^2}+\frac{\partial^2 f}{\partial t^2}\left(\frac{\partial t}{\partial x}\right)^2+\frac{\partial f}{\partial t}\frac{\partial^2 t}{\partial x^2}$

See if you can finish