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# Calculus Problem

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If $$y=arctan(cos(x))$$, then $$dy/dx$$=?

Oct 11, 2020

#2
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There are formulas for this, but this is how I would do it:

$$y=arctan(m),\quad\;\; m=cos(x)\\ tan(y)=m\\ \frac{dm}{dy}=sec^2(y)\qquad \frac{dm}{dx}=-sin(x)\\~\\ \frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}\\ \frac{dy}{dx}=\frac{1}{sec^2y}\cdot -sin(x)\\ \frac{dy}{dx}=\frac{-sin(x)}{sec^2(arctan(cos(x))}$$

Again there are formulas but I used a series of right angled triangles to simplify the denominator.

And I came up with  a denominator of      $$1+ cos^2(x)$$

The problem with my method is that I will only guarantee if for angles in the first quadrant.

Since it is squared it probably does not matter.

(that is assuming I made no stupid errors)

So now I have.

$$\frac{dy}{dx}=\frac{-sin(x)}{1+cos^2(x)}\\ \text{Wolfram Alpha gives the asnwer as} \\ \frac{dy}{dx}=\frac{-2sin(x)}{3+cos(2x)}\\~\\ BUT\\ \text{These two answers are rearrangements of the same thing.}$$

LaTex:

tan(y)=m\\
\frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}\\
\frac{dy}{dx}=\frac{1}{sec^2y}\cdot -sin(x)\\
\frac{dy}{dx}=\frac{-sin(x)}{sec^2(arctan(cos(x))}

Oct 11, 2020
edited by Melody  Oct 11, 2020
#3
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Here is a photo to show you the triangles I was talking about. Melody  Oct 11, 2020
#4
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Thank you so much for your detailed answer! This really really helps.

RandomUser  Oct 11, 2020