Question

prove the derivative of cos^-1(x) is -1 / radical(1-x^2)?

Answer 1

Let y = arccos(x). Then x = cos(y)

Now differente both sides ofthis last relation and use implicit differentiation.

\[\frac{d \ }{dx} x = \frac{d \ }{dx} \cos y = \frac{dy}{dx} \frac{d \ }{dy} \cos y\]

Answer 2

Now the left hand side, LHS = 1

and the right RHS = dy/dx . (-sin y) = dy/dx . - sqrt(1-x^2)

Answer 3

Hence
\[1 = \frac{dy}{dx} .( - \sqrt{1-x^2}) \implies \ \frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}}\]

Answer 4

you can use implicit diff, or the chain rule
\[x=\cos(\arccos(x))\]
\[1=-\sin(\arccos(x))\times \arccos'(x)\]
\[\arccos'(x)=-\frac{1}{\sin(\arccos(x))}\]
and then you have to think back to trig as to how to find
\[\sin(\arccos(x))=\sqrt{1-x^2}\] although that should be easy