Answer:
[tex]arccot(x)'=-\frac{1}{1+x^2}[/tex]
Step-by-step explanation:
Use implicit differentiation:
[tex]y=arccot(x)[/tex]
[tex]cot(y)=x[/tex]
[tex]\frac{1}{tan(y)}=x[/tex]
[tex]\frac{cos(y)}{sin(y)}=x[/tex]
[tex]\frac{dy}{dx}(\frac{cos(y)}{sin(y)})=\frac{dy}{dx}(x)[/tex]
[tex]\frac{sin(y)(-sin(y))-cos(y)cos(y)}{sin^2(y)}*\frac{dy}{dx}=1[/tex] (Quotient Rule: [tex]\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}[/tex] )
[tex]\frac{-sin^2(y)-cos^2(y)}{sin^2(y)}*\frac{dy}{dx}=1[/tex]
[tex]\frac{-(sin^2(y)+cos^2(y))}{sin^2(y)}*\frac{dy}{dx}=1[/tex]
[tex]\frac{-1}{sin^2(y)}*\frac{dy}{dx}=1[/tex]
[tex]-csc^2(y)\frac{dy}{dx}=1[/tex]
[tex]\frac{dy}{dx}=\frac{1}{-csc^2(y)}[/tex]
[tex]\frac{dy}{dx}=-sin^2(y)[/tex]
[tex]\frac{dy}{dx}=-sin^2(arccot(x))[/tex]
See the attached picture to understand how to evaluate the mixed composition of trig functions using a right triangle.
Therefore, the derivative of arccot(x) is [tex]-\frac{1}{\sqrt{1+x^2}}[/tex].
