if f(x)=4x arccos(3x+3) - sqrt(11-4x^2), what is f'(x)

Answer 1

if f(x)=4x arccos(3x+3) - sqrt(11-4x^2), what is f'(x)

Chain rule: df/dx = df/du * du/dx

Product rule: d( f(x).g(x))/dx = df/dx g + f dg/dx

derivative of Arccos x is little uncommon but its: -1/ sqrt(1-x2)

Starting from the second term we apply chain rule directly using u= 11- 4x2 and sqrt(u)

Thus: d ( sqrt(11-4x2)/dx = d sqrt(u)/ du * d(11-4x2)/dx = 1/2 * 1/sqrt(u) * (-8x) = -4x / sqrt(11-4x2) (a)

The chain rule for the arccos (3x+3) , which u = 3x+3 yields:

d (arccos u)/du * du/dx = -1/sqrt[ 1-(3x+3)^2] * 3 (b)

To finish, apply the product rule on first term and gather the results from (a) and (b)

f'(x) = 4*arccos(3x+3) - 4x *3/sqrt[ 1-(3x+3)^2] + -4x / sqrt(11-4x2)

Answer 2

f(x) = 4xArccos(3x+3) - √ (11 - 4x²)

f'(x) = 4Arccos(3x+3) + 4x[ -1 / √ (1 - (3x+3)²)(3)] - (1/2)(11 - 4x²)^-1/2 (-8x)

= 4Arccos(3x+3) - 12x / √ (-9x² - 18x - 8) + 4x / √(11 - 4x²)