[SOLVED] Derivative of an inverse trigonometric function

• Nov 19th 2009, 04:44 PM
yzobel
[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)))
• Nov 19th 2009, 05:03 PM
skeeter
Quote:

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

$\displaystyle \frac{d}{dt} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dt}$
• Nov 19th 2009, 05:14 PM
yzobel
Quote:

Originally Posted by skeeter
$\displaystyle \frac{d}{dt} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dt}$

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.
• Nov 19th 2009, 05:19 PM
RockHard
Isn't $\displaystyle {tan^-1(x)}$ just $\displaystyle cot(x)$?
• Nov 19th 2009, 05:23 PM
lvleph
Quote:

Originally Posted by RockHard
Isn't $\displaystyle {tan^-1(x)}$ just $\displaystyle cot(x)$?

No, $\displaystyle \tan^{-1} x = \arctan x$
• Nov 19th 2009, 06:07 PM
RockHard
Yes, I know that...$\displaystyle tan^-1(x)$ can be also written as $\displaystyle \frac{1}{tan(x)}$
which by trigonmetric identities $\displaystyle cot(x) = \frac{1}{tan(x)}$, I just did not want to state something and be wrong.

Table of Trigonometric Identities

Also it is whatever choice maybe the simplest because $\displaystyle \int cot(x)$ can be written as $\displaystyle \int\frac{cos(x)}{sin(x)}$ and from your problem above

$\displaystyle \int\frac{cos(x)}{sin(x)\sqrt{2-t^2}}$

And from here you can just use a trig. substitution using the identity, that fits the description of what's under the square root.
$\displaystyle sin^2(x)+cos^2(x) = 1$
• Nov 19th 2009, 06:12 PM
yzobel
Quote:

Originally Posted by RockHard
Yes, I know that...$\displaystyle tan^-1(x)$ can be also written as $\displaystyle \frac{1}{tan(x)}$
which by trigonmetric identities $\displaystyle cot(x) = \frac{1}{tan(x)}$, 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
• Nov 19th 2009, 06:52 PM
RockHard
Ah, simply overlooked, it my mistake good sir.
• Nov 20th 2009, 01:16 AM
mr fantastic
Quote:

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):

$\displaystyle \frac{1}{1 + u^2} = \frac{2 - t^2}{2}$

$\displaystyle \frac{du}{dt} = \frac{2}{(2 - t^2) \sqrt{2 - t^2}}$.

And thus the final answer for $\displaystyle \frac{dg}{dt}$ is simple.