Hi everyone, which way should I approach this one....
Differentiate with respect to x
$\displaystyle Y = 3 arccot 2$
Well, you can either just accept that
$\displaystyle \displaystyle\frac{d}{dx}\left[\cot^{-1}(x)\right]=\frac{-1}{1+x^2}$
or
If $\displaystyle y=\cot^{-1}(x)$, then $\displaystyle x=\cot(y)$
Now derive $\displaystyle x=\cot(y)$.
$\displaystyle \displaystyle 1=-\csc^2(y)\frac{dy}{dx}$
$\displaystyle \displaystyle\frac{dy}{dx}=-\frac{1}{\csc^2(y)}$
$\displaystyle \csc=\sqrt{1+\cot^2}=\sqrt{1+x^2}$
$\displaystyle \displaystyle\frac{dy}{dx}=-\frac{1}{1+x^2}$
$\displaystyle arccot(x)=cot^{-1}(x)$
$\displaystyle \displaystyle\frac{d}{dx}cot^{-1}(x)=-\frac{1}{1+x^2}$
$\displaystyle \displaystyle\frac{d}{dx}cot^{-1}(u)=-\frac{1}{1+u^2}\;\frac{du}{dx}$ from the chain rule
which gives
$\displaystyle \displaystyle\frac{d}{dx}cot^{-1}(2x)=?$
Using the notation "t"
$\displaystyle \displaystyle\frac{d}{dt}cot^{-1}(2t)=-\frac{1}{1+(2t)^2}\;\frac{d}{dt}(2t)$