Yes, but it's not as simple as saying yes:

**Written:** #f(x)=(x/(1-x))^3#, this function is the cube of a simpler function. We'll need the power rule and the chain rule.

The derivative of a cube is: #d/(dx)(u^3)=3u^2*(du)/(dx)#.

In this problem, #u= x/(1-x)#, so you'll need the quotient rule to find #(du)/(dx)#

#f'(x)=3(x/(1-x))^2 [(1(1-x)-x(-1))/(1-x)^2]#

#=3(x/(1-x))^2 [1/(1-x)^2]=(3x^2)/(1-x)^4#

**If you rewrite** the function as: #f(x)=x^3/(1-x)^3# You'll start with the quotient rule, and you'll need the power rule and chain rule when you differentiate the denominator.

#f'(x)=(3x^2*(1-x)^3- (x^3 )*3(1-x)^2(-1))/((1-x)^3)^2#

#f'(x)=(3x^2(1-x)^3+3 (x^3 )(1-x)^2)/(1-x)^6#

#f'(x)=(3x^2(1-x)^2[(1-x)+ x])/(1-x)^6=(3x^2)/(1-x)^4#