How do you differentiate #f(x)=sqrt(((3x^2-x^3)/(x-3x^2))# using the chain rule?

1 Answer
Dec 19, 2016

The first step is always to simplify the function as much as possible before differentiating.

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

#f(x) = sqrt((x(3- x))/(1 - 3x))#

Let's rewrite as #f(x) = sqrt(square)# where #square = (-x^2 + 3x)/(1 - 3x)#. Before applying the chain rule on #f(x)#, we need to apply the quotient rule to find #(d square)/(dx)#.

#d/dx square = ((-2x + 3)(1 - 3x) - (-3(-x^2 + 3x)))/(1 - 3x)^2#

#d/dx square = (-2x + 3 + 6x^2 - 9x - 3x^2 + 9x)/(1 - 3x)^2#

#d/dx square = (3x^2 - 2x + 3)/(1 - 3x)^2#

Now, use the chain rule to find the derivative of #f(x)#.

Let #y = sqrt(u) = u^(1/2)# and let #u = square#. Then #y' = 1/(2u^(1/2))# and #u' = (3x^2 - 2x + 3)/(1 - 3x)^2# (as derived above).

#y' = 1/(2u^(1/2)) xx (3x^2 - 2x + 3)/(1 - 3x)^2#

#y' = (3x^2 - 2x + 3)/(2(1 - 3x)^2sqrt((x(3 - x))/(1 - 3x)))#

Hopefully this helps!