Question

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

Answer 1

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

Explanation:

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.

By the power rule:

`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!