Question

tan^(-1)(3t)/t^(2)+1 I understand this for the most part I just keep screwing up applying the chain rule to tan^(-1)(3t)

Answer 1

You want to derive\[{\tan^{-1}(3t) \over {t^2+1}}\]Correct?

Answer 2

yes

Answer 3

Recall that the derivative of \[\tan^{-1}(x) ={1\over x^2+1}\]Then, since you have a \(3t\) instead of \(x\), \[{d \over dt}\quad \tan^{-1}(3t)={1 \over {(3t)^2 +1}} \cdot 3\]Can you solve the rest?

Answer 4

you need the quotient rule to differentiate this

Answer 5

got it!! thank you!!

Answer 6

You're welcome.

Answer 7

((t^2 + 1)x(3/)9t^2 + 1) - tan^(-1)(3t)x(2t))/(t^2 + 1)^2