Thread: [SOLVED] Derivative of an inverse trigonometric function

I need help deriving this function:

g(t)=tan^-1 (t/ sqrt(2 - t^2))

Any help would be appreciated! Thanks


What I got when I tried deriving it:
1/(1+(t/sqrt(2-t^2))^2) x (sqrt(2-t^2)+(t^2/(sqrt(2-t^2)))/(sqrt(2-t^2))
and eventually ended up with:
2/((sqrt(2-t^2))^3 + t^3(sqrt(2-t^2)))

Originally Posted by yzobel

I need help deriving this function:

g(t)=tan^-1 (t/ sqrt(2 - t^2))

Any help would be appreciated! Thanks

Originally Posted by skeeter

Thank you, but I already tried that before and came up with a long, complicated answer. My teacher mentioned that g should be simplified beforehand but I'm not quite sure how to do that.

Isn't just ?

Originally Posted by RockHard

Isn't just ?

No,

Yes, I know that... can be also written as
which by trigonmetric identities , I just did not want to state something and be wrong.

Table of Trigonometric Identities

Also it is whatever choice maybe the simplest because can be written as and from your problem above



And from here you can just use a trig. substitution using the identity, that fits the description of what's under the square root.

Originally Posted by RockHard

Yes, I know that... can be also written as
which by trigonmetric identities , I just did not want to state something and be wrong.

Table of Trigonometric Identities

They're not the same.

tan^-1(x)=arctanx

(tanx)^-1 =cotx

Ah, simply overlooked, it my mistake good sir.

Originally Posted by yzobel

Thank you, but I already tried that before and came up with a long, complicated answer. My teacher mentioned that g should be simplified beforehand but I'm not quite sure how to do that.

The answer is not long and is not complicated.

After a small amount of algebra (at this level it should be seen as small):



.

And thus the final answer for is simple.