#1**+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**+10 **

Best Answer

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**0 **

Here's a link that will help you here.......http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx

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