Question

A little help please? Inverse trig. functions?

Answer 1

How would I find the derivative of \[f(x)= \sin( arc tanx)\]

Answer 2

I get that there are given derivatives for sin inverse, cos inverse, etc. But what do I do with the arc tan x?

Answer 3

you can use chain rule

Answer 4

$$\Large {
f(x)= \sin( \arctan(x))
\\ f ' (x) = \cos(\arctan(x) ) \cdot (\arctan(x)) '
}
$$

Answer 5

do you know the derivative of arctangent ?

Answer 6

\[y=\tan^{-1}(x)\]\[x=\tan(y)\]differentiate,\[1=\frac{dy}{dx}\sec^2(y)\]\[1=\frac{dy}{dx}(\tan^2(y)+1)\]\[\frac{dy}{dx}=1/(\tan^2(y)+1)\]\[\frac{dy}{dx}=1/(x^2+1)\]

Answer 7

just showing why

Answer 8

Okay so if I inputed what tan^-1x is equal to, would the chain rule give me \[\cos (arc \tan x) (\frac{ 1 }{ 1+x^2 })\]?

Answer 9

yes, that is right.

Answer 10

multiplying the derivative of the entire argument times its chain rule

Answer 11

Where do I go from there?

Answer 12

that is it, you got the answer, and do not need anything else

Answer 13

Okay, got it. Thank you

Answer 14

\[\frac{ dy }{ dx }~\left[{\rm Sin}({\rm ArcTan}~(x)~)\right]~=~{\rm Cos}({\rm ArcTan}~(x)~) \times \left(\frac{1}{x^2+1}\right)\]

Answer 15

bye

Answer 16

You could also convert to algebraic, first, if you like.

\(\cos(atan(x)) = \dfrac{1}{\sqrt{1+x^{2}}}\)