# Thread: Need refresher on derivatives

1. ## 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.

2. Hi

You are right but you made a useless complication
$u(x,y,z)=A\:\sin(ax)\:\cos(by)\:\sinh(cz)$

As you said you must consider $\cos(by)\:\sinh(cz)$ as constant

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

You can consider $u(x,y,z)=B\:\sin(ax)$ where $B=A\:\cos(by)\:\sinh(cz)$

Therefore the partial derivative of u with respect to x is

$\frac{\partial u(x,y,z)}{\partial x}=Ba\:\cos(ax)$

and the second derivative is

$\frac{\partial^2 u(x,y,z)}{\partial x^2}=-Ba^2\:\sin(ax) = -Aa^2\:\sin(ax)\:\cos(by)\:\sinh(cz)$

You do not need to use the product rule

3. 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!