We actually just studied this last semester, so this might be helpful. The formula for the chain rule is: (u)^n --> n(u)^(n-1) (u')

What this means is that you start it like you would start a normal derivative: you take the exponent and move it to the front of the expression, and then subtract one from (n) and that becomes the new exponent. So, for example, if I were to take the equation y=(x-5)^3, I would start this problem by moving the 3 to the front, so it would look like 3(x-5), and then I would subtract 1 from 3, and get 2 as my new exponent, which would look like 3(x-5)^2. After that, you take the derivative of (u), which in this case would just be 1 (derivative of 5=0, derivative of x=1). So you'd multiply the entire thing by 1, 3(x-5)^2 (1).

Step-by-step:

y=(x-5)^3

[move exponent] 3(x-5)

[subtract and make new exponent] 3(x-5)^2

[multiply by drv. of inside] 3(x-5)^2 (1)

Y'=3(x-5)^2 (1)

I hope this helps. If you need more help, feel free to PM me and I'll help where I can.