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

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Answer 1 · 2016-06-07T15:17:09

The chain rule states that #dy/dx = dy/(du) xx (du)/dx#.

First, though we have to break this function up into less complex functions.

Let #y = sqrt(u) = u^(1/2)#

Let #u = (3x)/(2x - 3)#

We're going to have to differentiate both functions. Let's start with the easiest: y.

#y' = 1/2u^(1/2 - 1)#

#y' = 1/(2u^(1/2)#

Now to #u#. For this function, we must differentiate using the quotient rule.

Let #u(x) = (g(x))/(h(x))#. The quotient rule states that #u'(x) = (g'(x) xx h(x) - g(x) xx (h'(x)))/(h(x))^2#.

In our function, let #g(x) = 3x# and #h(x) = 2x - 3#.

By the power rule #g'(x) = 3# and #h'(x) = 2#

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

#u'(x) = (6x - 9 - 6x)/(4x^2 - 12x + 9)#

#u'(x) = -9/(4x^2 - 12x + 9)#

Now, substituting:

#f'(x) = 1/(2sqrt((3x)/(2x - 3))) xx -9/(4x^2 - 12x + 9#

Hopefully this helps!