Question

I have a web-assign problem that I have the answer to by examining the solutions from alternate, but related questions. I have it right, but in case this problem shows up, I would like to know how to solve it.
find derivative of y=tan^-1(z/(8-z^2)) the solution is (8+z^2)/((8-z^2)^2+z^2), but I come up with (8+z^2)/((8-z^2)^2+z(8-z)) and I don't know what I did wrong. I used the chain rule with tan^-1 having the derivative of 1/(1+g()^2) where g()=(z/(8-z^2)) and then multiplied by g'() using the quotient rule solves to (1*(8-z^2)-z(8-z^2))/(8xz^2)^2. Am I starting wrong?

Answer 1

y = tan^-1(z/(8-z^2))
let u = z/ (8-z^2)
use quotient rule:
du/dz = (8-z^2) - z(-2z)
---------------
(8-z^2)^2

= 8 + z^2
-------
( 8 - z^2)^2
y = tan^-1z
dy/du = 1 / 1 + u^2

dy/dz = dy/du * du/dz
= 1 8 + z^2
-- * --------
1 + (z^2/(8-z^2)^2 ( 8 - z^2)^2

= (8-z^2)^2 * (8+z^2)
-------- -------
(8-z^2)^2 + z^2 (8-z^2)^2

= (8+z^2)
-----------
(8-z^2)^2 + z^2

Answer 2

i find it easier to use the 'dy/dx' notation when doing more complicated derivatives like this

Answer 3

Thank you, I found the mistake, I forgot the square on the derivative of the inverse tangent. Your steps lead me to find the mistake, thank you.