Thread: Need refresher on derivatives

I am taking a partial differential class this fall, unfortunately it has been about 12 years since I have had calculus and I have been trying to refresh my memory, but there are a couple of things I just need verification for a few things.

Given:

u(x,y,z)=A*sin(ax)*cos(by)*sinh(cz)

A,a,b,c are constants.

Find the second derivative with respect to x:

If I remember correctly, when when taking a derivative with respect to one variable the other variables are treated as constants. In this case when you apply the product rule the first derivative would look like this:

du/dx=0*sin(ax)*cos(by)*sinh(cz)+A*(a)cos(ax)*cos(by) *sinh(cz)+A*sin(ax)*0*sinh(cz)+A*sin(ax)*cos(by)*0

du/dx=A*(a)cos(ax)*cos(by)*sinh(cz)

It would then follow that the second derivative would look like:

d^2u/dx^2=0*(a)cos(ax)*cos(by)*sinh(cz)+A*(a^2)(-sin(ax))*cos(by)*sinh(cz)+A*(a)cos(ax)*0*sinh(cz)+ A*(a)cos(ax)*cos(by)*0

d^2u/dx^2=-A*(a^2)sin(ax)cos(by)sinh(cz)

so basically the second derivative with respect to x looks like:

u''=-A(a^2)sin(ax)cos(by)sinh(cz)

So can someone help me out by either confirming my solution or pointing out where I have made mistakes. I appreciate the help.

Hi

You are right but you made a useless complication


As you said you must consider as constant

The derivative of kf (with k constant and f function) is kf'

You can consider where

Therefore the partial derivative of u with respect to x is



and the second derivative is



You do not need to use the product rule

Thanks, I follow the logic for simplifying the problem, but until I restore my confidence in applying all the rules correctly I wanted to do everything the long way. I appreciate the help and the confirmation, that goes a long way to helping to restore my confidence!