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I'm working on an exact equation, and there's a derivative in there that is throwing me off. I can solve it one way, but when I try solving it with a different method, I seem to get the wrong answer and I can't figure out why. Here it is:

So let's say I need to find

$\frac{d}{dy}arctan (y)$

Here's what I did:

$\frac{d}{dy}arctan (y)=\frac{1}{tan'(arctan(y))}=\frac{1}{1+tan^2(arctan(y))}$

$\frac{d}{dy}arctan (y)=\frac{1}{1+y^2}$

Ok. But now I want to try to solve it another way. This time I use $tan(y)'=sec^2(y)$ instead of $1+tan^2(y)$, so I have

$\frac{d}{dy}arctan (y)=\frac{1}{sec^2(arctan(y))}$

I try to simplify the denominator:

$\frac{1}{sec^2(arctan(y))}=cos^2(arctan(y))=\frac{1}{(1+y^2)^2}$

I've managed to somewhere square my answer. How did I do that?

Korgan Rivera
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1 Answers1

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Note that $\cos\left(\arctan\left(y\right)\right)=\frac{1}{\sqrt{1+y^{2}}}$ so that $\cos^{2}\left(\arctan\left(y\right)\right)=\frac{1}{1+y^{2}}$.

Draw a right triangle with opposite and adjacent sides (to the angle $\arctan\left(y\right)$) of lengths $y$ and $1$, respectively. Then, the hypotenuse is $\sqrt{1+y^{2}}$. You just forgot to take the square root when calculating the cosine of the arc-tangent.

Brian
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