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# derivative of arctan(x/2)

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derivative of arctan(x/2)

Guest Sep 14, 2014

#1
+94202
+10

There is some formula that you are supposed to know but I can never remember it.

so I'll do it the long way.

$$\\let\quad \theta=atan(x/a)\\ tan\theta=\frac{x}{a}\\\\ x=a*tan\theta\\\\ \frac{dx}{d\theta}=a*sec^2\theta\\\\ \frac{d\theta}{dx}=\frac{cos^2\theta}{a}\\\\ \frac{d\theta}{dx}=\frac{\frac{a^2}{x^2+a^2}}{a}\\\\ \frac{d\theta}{dx}=\frac{a^2}{a(x^2+a^2)}\\\\ \frac{d\theta}{dx}=\frac{a}{a^2+x^2}\\\\$$

I expect it is more sensible to remember the formua  LOL

Melody  Sep 14, 2014
#1
+94202
+10

There is some formula that you are supposed to know but I can never remember it.

so I'll do it the long way.

$$\\let\quad \theta=atan(x/a)\\ tan\theta=\frac{x}{a}\\\\ x=a*tan\theta\\\\ \frac{dx}{d\theta}=a*sec^2\theta\\\\ \frac{d\theta}{dx}=\frac{cos^2\theta}{a}\\\\ \frac{d\theta}{dx}=\frac{\frac{a^2}{x^2+a^2}}{a}\\\\ \frac{d\theta}{dx}=\frac{a^2}{a(x^2+a^2)}\\\\ \frac{d\theta}{dx}=\frac{a}{a^2+x^2}\\\\$$

I expect it is more sensible to remember the formua  LOL

Melody  Sep 14, 2014
#2
+93025
0

Note that, if we let a = 1 in Melody's derivation, then her "formula" is exactly correct.....!!!!

Well done, Melody !!!

[ Note to the site users......it's always more satisifying when you can derive "formulas" on your own rather than relying on "tables"......you may not be exactly correct, but at least you might have a "feel" for how it was originally derived by the "pros" ]

CPhill  Sep 14, 2014