1. ## Derivative

differentiate ))

Hey I must say this is an awesome site!!

q1.) y = tanh ^ -1 (3x^5)

using chain rule.

u=3x^5 du/dx = 15x^4

y= tanh ^ -1 (u) dy/du = 1/1-u^2 . (u) >>> SO; 1/(1-3x^5)^2 . (u)

y' = (3x^5 / 1 - 9x^10) . (15x^4)

or should it just be

y' = 15x^4 / 1-9x^10

2. Originally Posted by Jon123
differentiate ))

Hey I must say this is an awesome site!!

q1.) y = tanh ^ -1 (3x^5)

using chain rule.

u=3x^5 du/dx = 15x^4

y= tanh ^ -1 (u) dy/du = 1/1-u^2 . (u) >>> SO; 1/(1-3x^5)^2 . (u)

y' = (3x^5 / 1 - 9x^10) . (15x^4)

or should it just be

y' = 15x^4 / 1-9x^10
The second answer is correct. I'm not sure why you are including the 3x^5 in the numerator of the first.

-Dan

Edit: Oh your second answer was correct, but a little matter of typing. This line has problems:
dy/du = 1/1-u^2 . (u) >>> SO; 1/(1-3x^5)^2 . (u)

First dy/du = 1/(1 - u^2) (watch the parenthesis!), so upon subbing in u = 3x^5 you get dy/dx = 1/(1 - (3x^5)^2) * du/dx. Note the form of the denominator. Also note that it is du/dx that shows up here, not u. The full chain rule is
$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$