Question

find the derivative of 2x arccos(x)-2square root 1-x^2, i know that the answer is 2arccos x, help please

Answer 1

Where are you stuck?

Answer 2

i see formula for the derivative, and i am not sure what to plug or combine like terms

Answer 3

Let @Kainui help, Kainui ha??

Answer 4

would you use productive rule?

Answer 5

I'm just watching good luck @OOOPS !

Answer 6

@MrGrimm show me your work, please

Answer 7

Of course we have to use product rule.

Answer 8

this may be helpful

http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Derivatives_of_inverse_trigonometric_functions

and yes you will be using the product rule as well as the chain rule

Answer 9

lets look at first term, using product rule

\[\rightarrow (2x)' \cos^{-1} x + 2x(\cos^{-1} x)'\]
\[= 2 \cos^{-1} x - \frac{2x}{\sqrt{1-x^2}}\]

Answer 10

\[(-2\sqrt{1-x ^{2}})(2\arccos x)+(2xarccos x)(2\sqrt{1-x ^{2}})\]

Answer 11

thats what i thought it would be

Answer 12

ok for 2nd term, we use chain rule

\[\rightarrow \frac{-2 (-2x)}{2\sqrt{1-x^2}}\]
because derivative of "1-x^2" is -2x
simplifying to
\[2 \cos^{-1} x -\frac{2x}{\sqrt{1-x^2}}+\frac{2x}{\sqrt{1-x^2}}\]
\[= 2 \cos^{-1} x\]

Answer 13

ok thats more clear thank you, was confused on what to take the derivative

Answer 14

yw