Question

I need help solving this problem.
`4x cos ^(-1) (2x+4)- \sqrt{3-3 x^(2) },` find f'(x). My original which is wrong, is included.

Answer 1

Given:

`f(x)=4xcos^(-1)(2x+4)-√(3-3x^2)`

Using

`(d)/(dx)cos^(-1)(x)=-(1)/(√(1-x^2))`
we derive

`(d)/(dx)4xcos^(-1)(2x+4)`

`=4cos^(-1)(2x+4)-(8x)/(√(1-(2x+4)^2))`

Similarly, using

`(d)/(dx)√(x)=(1)/(2√(x))`
we derive

`(d)/(dx)(-√(3-3x^2))`

`=(3x)/(√(3-3x^2))`

Therefore, the derivative is

`f'(x)=(d)/(dx)(4xcos^(-1)(2x+4)-√(3-3x^2))`

`=4cos^(-1)(2x+4)-(8x)/(√(1-(2x+4)^2))+(3x)/(√(3-3x^2))`

Answer 2

I believe it's as follows:


`f'(x) = 4(acos(2x+4)- (2x)/( √(-4x^2-16x-15)))+ (3x)/( √(3-3x^2))`