1. ## Inverse Trig

Not sure what i'm getting wrong. Here is the question.

I used the Pythagorean theorem, therefore arc(x) = opp/adj, hence my hypotenuse was x+1 since (x^2+1)^1/2, Now I assume that Sec^2= Hyp^2/Adj ^2. The result just seems wrong and it doesn't help me prove arctan, thx fox the help.

2. Since you didn't give the previous result we can only GUESS at what you need?
Here is my guess first note that

$\displaystyle \displaystyle \tan\left( \tan^{-1}(x)\right)=x$

Now if we take the derivative using the chain rule on the left hand side we get

$\displaystyle \displaystyle \frac{d}{dx}\tan\left( \tan^{-1}(x)\right)=\frac{d}{dx}x$

$\displaystyle \displaystyle \sec^2\left( \tan^{-1}(x)\right)\left(\frac{d}{dx} \tan^{-1}(x)\right)=1$

Now solving for the derivative we want gives

$\displaystyle \displaystyle \frac{d}{dx}\tan^{-1}(x)=\frac{1}{\sec^2(\tan^{-1}(x))}=...$

Now just use what you found out in this part of the problem.

3. The previous result is actually referring to number 2. Why did u take the derivative to find the algebraic form? Thx for the help.

Originally Posted by TheEmptySet
Since you didn't give the previous result we can only GUESS at what you need?
Here is my guess first note that

$\displaystyle \displaystyle \tan\left( \tan^{-1}(x)\right)=x$

Now if we take the derivative using the chain rule on the left hand side we get

$\displaystyle \displaystyle \frac{d}{dx}\tan\left( \tan^{-1}(x)\right)=\frac{d}{dx}x$

$\displaystyle \displaystyle \sec^2\left( \tan^{-1}(x)\right)\left(\frac{d}{dx} \tan^{-1}(x)\right)=1$

Now solving for the derivative we want gives

$\displaystyle \displaystyle \frac{d}{dx}\tan^{-1}(x)=\frac{1}{\sec^2(\tan^{-1}(x))}=...$

Now just use what you found out in this part of the problem.

4. Originally Posted by melvis
The previous result is actually referring to number 2. Why did u take the derivative to find the algebraic form? Thx for the help.
I didn't take the derivative to find the algebraic form. What you found in the OP is the correct algebraic form($\displaystyle \sec^2(\tan^{-1}(x))=x^2+1$), If you plug that into what I did you will get the answer that you are looking for.