What is the derivative of #(4-x)/x^(1/3)(6-x)^(2/3)#?

Question

Paisley Johnson · Accepted Answer

Ew. If we assume that function is #f(x)#

#f'(x) = (4x^2−16x−24)/(3*(6-x)^(1/3)x^(4/3)# Break up the equation and tackle it piece by piece. I'll let you do the calculations but i'll simplify how it looks: Let #a=(4-x) # Let #b=(x^(1/3)) # Let #c=(6-x)^(2/3) # #f'(a/b) = [[d/dx(a) * (b)] - [d/dx(b) * (a)]]/b^2 # Then use the product rule to finish it. #f'(x) = f'(a/b) * (c) + (a/b) * d/dx(c) #

Emily Ammerman · Answer

undefined To find the derivative of the given function, apply the product rule and chain rule.

The derivative of the function is:

$$ \frac{d}{dx}\left(\frac{4-x}{x^{1/3}(6-x)^{2/3}}\right) = \frac{-1}{3x^{4/3}(6-x)^{2/3}} + \frac{2(4-x)}{3x^{4/3}(6-x)^{5/3}} - \frac{(4-x)}{x^{4/3}(6-x)^{2/3}} $$

$$ = \frac{-1}{3x^{4/3}(6-x)^{2/3}} + \frac{8-2x}{3x^{4/3}(6-x)^{5/3}} - \frac{(4-x)}{x^{4/3}(6-x)^{2/3}} $$

$$ = \frac{-3(4-x) + (8-2x)}{3x^{4/3}(6-x)^{5/3}} $$

$$ = \frac{-12 + 3x + 8 - 2x}{3x^{4/3}(6-x)^{5/3}} $$

$$ = \frac{x-4}{3x^{4/3}(6-x)^{5/3}} $$