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:
$f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}$

Using
$\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}$
we derive
$\frac{d}{dx}4xcos^{-1}(2x+4)$
$=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}$

Similarly, using
$\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$
we derive
$\frac{d}{dx}(-\sqrt{3-3x^2})$
$=\frac{3x}{\sqrt{3-3x^2}}$

Therefore, the derivative is
$f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})$
$=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}$

Answer 2

I believe it's as follows:

$f'(x) = 4(acos(2x+4)- \frac{2x}{ \sqrt{-4x^2-16x-15}})+ \frac{3x}{ \sqrt{3-3x^2}}$