Question

How do you find the derivative for `y= (2x) / (sqrt(x - 1))`?

Answer 1

You can use two approaches to differentiate this function.

Explanation:

A very important thing to notice about this function is that you can write both as a quotient of two other functions, let's say `f(x)` and `g(x)`, and a product of two functions, `f(x)` and `h(x)`.

This means that you can approach it in two different ways, either by using the product rule or by using the quotient rule.

The chain rule will be needed in both cases, though.

To keep the answer from becoming too long, I will only show you one of the two approaches.

So, your function can be written as

`y = f(x)/g(x) = (2x)/sqrt(x-1)`

The quotient rule allows you to differentiate this function by using

`color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2)`

This means that you have

`d/dx(y) = ([d/dx(2x)] * sqrt(x-1) - (2x) * d/dx(sqrt(x-1)))/(sqrt(x-1))^2`

You can calculate `d/dx(sqrt(x-1))` by using the chain rule for

`d/dx(u^(1/2))`, with `u = x-1`

This will get you

`d/dx(u^(1/2)) = d/(du)u^(1/2) * d/dx(x-1)`

`d/dx(u^(1/2)) = 1/2 * u^(-1/2) = 1/2 * 1/(sqrt(x-1))`

Your target derivative will now be

`d/dx(y) = (2 * sqrt(x-1) - color(red)(cancel(color(black)(2))) * x * 1/color(red)(cancel(color(black)(2))) * 1/(sqrt(x-1)))/(x-1)`

You can simplify this by using fractional exponents

`d/dx(y) = (2 * (x-1)^(1/2) - x * (x-1)^(-1/2))/(x-1)`

`d/dx(y) = [(x-1)^(-1/2) * [2 * (x-1) - x]]/(x-1)`

`d/dx(y) = color(green)((x-2) * (x-1)^(-3/2))`

The starting point for the product rule approach would have been

`d/dx(y) = [d/dx(2x)] * (x-1)^(-1/2) * 2x * d/dx(x-1)^(-1/2)`