# Derivatives, and a little bit of linear approxim

## Homework Statement

I have to find the deriv
of ##f(x)=xArctan(x^{3})##

I just need an explanation of how the arctan works...
So I understand the rest, but just the deriv of arctan itself is confusing for me.
So the derivative of ##arctan(x)## just in general is ##\frac{1}{1+x^{2}}##

But if it's of ##arctan(x^{3})##
Then I'm not sure what is right

So on my test I was pretty sure that when I derived arctan, that the x^2 factor meant (x)^2, so I would have to put in ##x^3##, and then I'd get ##\frac{1}{1+x^{6}}##
(That's x to the 6th, because you'd mult the exps)

But.... my test says I forgot to chain rule the ##x^{3}## that's inside of arctan(), but why did the teacher write (.3x^2) as the ##g'(x)##?

But I also wrote ##\frac{1}{1+3x^{4}}##, which wasn't marked wrong as the deriv of ##arctan(x^{3})##, so.. I'm confused.

SteamKing
Staff Emeritus
Homework Helper
If y = tan (x), then x = arctan (y), that is, arctan is the inverse function of tan.

For d/dx [arctan (x^3)], you should make the substitution u = x^3,

then d/dx [arctan (x^3)] = d/du [arctan (u)] * du/dx, by the chain rule.

The derivative of arctan u = 1/(1+u^2) [from the formula] and I'm sure you can calculate du/dx, when u = x^3

So.. wait I'm sorry, so then what's the formula?
On your last line I'm not so clear what you mean, sorry.

SteamKing
Staff Emeritus