Question

Prove the formula for (d/dx)(cos−1(x)) by the same method as for (d/dx)(sin−1(x)). Let y = cos−1(x). Then cos(y) = and 0 ≤ y ≤ π ⇒ −sin(y) dy dx = 1 ⇒ dy dx = − 1 sin(y) = − 1 1 − cos2 = − 1 1 − x2 .

Answer 1

Answer:

$\frac{d\,arccos(x)}{dx} =-\frac{1}{\sqrt{1-x^2} }$

See the proof below

Step-by-step explanation:

if  $y=arccos(x)$, then :  $cos(y)=cos(arccos(x))=x$

We can then apply the derivative to both sides of this last equation and remember to use the chain rule as we encounter the variable "y":

$cos(y)=x\\\frac{d}{dx} cos(y)=\frac{d}{dx} x\\-sin(y)\,\frac{dy}{dx} =1\\\frac{dy}{dx} =-\frac{1}{sin(y)} \\\frac{dy}{dx} =-\frac{1}{\sqrt{1-cos^2(y)} }\\\frac{dy}{dx} =-\frac{1}{\sqrt{1-x^2} }\\\frac{d\,arccos(x)}{dx} =-\frac{1}{\sqrt{1-x^2} }$