Question

Use the inverse function-inverse cofunction

Answer 1

Answer:

$arccot(x)'=-\frac{1}{1+x^2}$

Step-by-step explanation:

Use implicit differentiation:

$y=arccot(x)$

$cot(y)=x$

$\frac{1}{tan(y)}=x$

$\frac{cos(y)}{sin(y)}=x$

$\frac{dy}{dx}(\frac{cos(y)}{sin(y)})=\frac{dy}{dx}(x)$

$\frac{sin(y)(-sin(y))-cos(y)cos(y)}{sin^2(y)}*\frac{dy}{dx}=1$ (Quotient Rule: $\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ )

$\frac{-sin^2(y)-cos^2(y)}{sin^2(y)}*\frac{dy}{dx}=1$

$\frac{-(sin^2(y)+cos^2(y))}{sin^2(y)}*\frac{dy}{dx}=1$

$\frac{-1}{sin^2(y)}*\frac{dy}{dx}=1$

$-csc^2(y)\frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{-csc^2(y)}$

$\frac{dy}{dx}=-sin^2(y)$

$\frac{dy}{dx}=-sin^2(arccot(x))$

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 $-\frac{1}{\sqrt{1+x^2}}$.