Question

Quick derivative question:

Answer 1

What is \(\LARGE \frac{d}{dx}~(arctan~4x)\)?

Answer 2

If my memory is correct it would be

\[\large \frac{4}{1 + 16x^2}\]

Basically by using the chain rule

Answer 3

Correct.

Answer 4

\[\large \frac{d}{dx} tan^{-1}(x) = \frac{1}{1 + u^2}\]
\[\large \frac{d}{dx} 4x = 4\]

inside times outside = \(\Large \frac{4}{1 + 16x^2}\)

Answer 5

Alright, thank you guys :)

Answer 6

dx/d = something *shrugs*