Finding Derivative of V= ay/(b2 + y2) - Help Needed

  • Thread starter Thread starter Kali8972
  • Start date Start date
  • Tags Tags
    Integral Partial
Kali8972
Messages
14
Reaction score
0
Partial integral?

Can someone help me figure out what this means? It's been forever since I've had any math classes.

V= ay/(b2 + y2)

Find the derivative in terms of a,b, and y

Thanks so much!
 
Physics news on Phys.org
Kali8972 said:
Can someone help me figure out what this means? It's been forever since I've had any math classes.
V= ay/(b2 + y2)
Find the derivative in terms of a,b, and y
Thanks so much!
Do you know what a derivative is? Nobody here is going to just do the problem for you, but if you show what you're having trouble with we'll try to help :smile:
 
yes... For example if you had 2x^3 the derivative would be 6x^2... When i do the derivative for the above equation I get:
a(b^2-y^2)/(b^2+y^2)^2
 
Use the "quotient rule": the derivative of \frac{f(y)}{g(y)} is
\frac{\frac{df}{dy}g(y)- f(y)\frac{dg}{dy}}{g^2}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top