Question

What is the derivative of `g(u) = ln(sqrt((3u+6)/(3u-6)))`?

Answer 1

`g^' = -2 * 1/((u+2)(u-2))`

Explanation:

First, notice that you can simplify the function o get

`g(u) = ln(sqrt((color(red)(cancel(color(black)(3)))(u+2))/(color(red)(cancel(color(black)(3)))(u-2)))) = ln(sqrt((u+2)/(u-2)))`

You can differentiate this function by using the chain rule twice and the quotient rule once.

More specifically, you'll need to use the chain rule for

• `lnt`, with `t = sqrt((u+2)/(u-2))`
• `sqrt(v)`, with `v = (u+2)/(u-2)`

So, start with the target derivative

`d/(du)(g) = d/(dt)lnt * d/(du)(t)`

`g^' = 1/t * d/(du)(sqrt((u+2)/(u-2)))`

`g^' = 1/t * d/(dv)sqrt(v) * d/(du)(v)`

`g^' = 1/t * 1/2 * 1/sqrt(v) * d/(du)((u+2)/(u-2))`

This is where the quotient rule comes in handy

`d/(du)((u+2)/(u-2)) = ([d/dx(u+2)] * (u-2) - (u+2) * d/dx(u-2))/(u-2)^2`

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

`d/(du)((u+2)/(u-2)) = (color(red)(cancel(color(black)(u))) - 2 - color(red)(cancel(color(black)(u))) - 2)/(u-2)^2`

`d/(du)((u+2)/(u-2)) = -4/(u-2)^2`

This means that the target derivative will be equal to

`g^' = 1/2 * 1/sqrt((u+2)/(u-2)) * 1/sqrt((u+2)/(u-2)) * (-4/(u-2)^2)`

`g^' = - 2 * (1/sqrt((u+2)/(u-2)))^2 * 1/(u-2)^2`

`g^' = -2 * color(red)(cancel(color(black)(u-2)))/(u+2) * 1/(u-2)^color(red)(cancel(color(black)(2)))`

`g^' = color(green)(-2 * 1/((u+2)(u-2)))`

Answer 2

`g'(u) = (-2)/((u+2)(u-2)) = (-2)/(u^2-4)`

Explanation:

Factor and reduce the innermost fraction, then use properties of logarithms to rewrite `g`

`g(u) = ln(sqrt((3u+6)/(3u-6)))= ln(sqrt((3(u+2))/(3(u-2))))`

`= ln(sqrt((u+2)/(u-2)))`

`= 1/2ln((u+2)/(u-2))`

`= 1/2[ln(u+2)-ln(u-2)]`

Now differentiate with respect to `u`:

`g'(u) = 1/2[1/(u+2)-1/(u-2)]`

`= 1/2[((u-2)-(u+2))/((u+2)(u-2))]`

`= 1/2[(-4)/((u+2)(u-2))]`

`= (-2)/((u+2)(u-2)) = (-2)/(u^2-4)`