find derivative of function. Simplify when possible.

y=xcos^-1x - square root 1-x^2

The answer is 1.





I'm lying. Do your own homework.

I agree with techsupport, although I am happy to give you help with your work, it helps if you ask for help, rather than just demanding at us.

Secondly, I don't understand your question, what does xcos^-1x mean?

Capuchin, I believe this person means xarccos(x).

Phil

so y = xarccos(x) - sqrt{1-x^2}

Finding \frac{dy}{dx}?

Hmmm let me have a think.

the hard part is the derivative of arccos(x). You can either show that for the domain defined by the open unit disc centered on the origin you have arccos(x) = integral from x to 1 of 1/(1 + x^2)^.5 and then use the fundamental thm of calculus (don't forget to switch the limits of integration!) or you can consider theta = arccosx so that d/dx(arccos(x)) = d(theta)/d(cos(theta)) = 1/(-sin(theta)) = -1/(1 - (cos(theta))^2)^.5 = -1/(1 - x^2)^.5.