Question

tan^-1(ln x^2) what's the derivative

Answer 1

chain rules ....

Answer 2

[tan^-1] '(a) * ln'(b) * (x^2)'

Answer 3

2x/1+ (ln x^2)^2 is what im getting

Answer 4

hmm

tan-1(a) = a'/(1+a^2) , but a = ln(b)
a' = b'/b

b'/b(1+(ln b)^2), but b=x^2
b' = 2x


2x/x^2(1+(ln x^2)^2)

2/x(1+ (ln x^2)^2)

Answer 5

you was close

Answer 6

Okay thanks I always have trouble with the chain rule I need to be more careful

Answer 7

yes, being careful is key, which is why when they get complicated looking i do like i did ... focus on the parts that are deriving and fill in as i go