Question

What is the derivative of `ln[sqrt((2+x^2)/(2-x^2))]`?

Answer 1

`y^' = (4x)/((2-x^2) * (2 + x^2))`

Explanation:

You're going to have to use a combination of derivative rules to differentiate this function.

Right from the start, it's clear that you'll use the chain rule twice, for

• `y = ln(u)`, with `u = sqrt((2+x^2)/(2-x^2))`
• `z = sqrt(u_1)`, with `u_1 = (2+x^2)/(2-x^2)`

At that point, you can use the quotient rule to complete the integration.

So, the derivative of `y` will be

`d/dx(y) = d/(du) * ln(u) * d/dx(underbrace(sqrt((2+x^2)/(2-x^2)))_(color(blue)(z)))`

Now focus on finding `d/dx(color(blue)(z))`, which will be equal to

`d/dx(sqrt(u_1)) = [d/(du_1)sqrt(u_1)] * d/dx(underbrace((2+x^2)/(2-x^2))_(color(orange)(w)))`

This time, use the quotient rule to find `d/dx(color(orange)(w))`

`d/dx(w) = ([d/dx(2+x^2)] * (2 - x^2) - (2 + x^2) * d/dx(2 - x^2))/(2-x^2)^2`

`w^' = [2x(2-x^2) - (2 + x^2) * (-2x)]/(2-x^2)^2`

`w^' = (4x - color(red)(cancel(color(black)(2x^3))) + 4x + color(red)(cancel(color(black)(2x^3))))/(2-x^2)^2`

`w^' = (8x)/(2-x^2)^2`

Plug this into the calculation of `d/dx(color(blue)(z))` to get

`d/dx(sqrt(u_1)) = 1/2u_1^(-1/2) * (8x)/(2 - x^2)^2`

`d/dx(sqrt((2+x^2)/(2-x^2))) = 1/2 * 1/(sqrt((2+x^2)/(2-x^2))) * (8x)/(2-x^2)^2`

Finally, plug this into the calculation for the target derivative to get

`d/dx(lnu) = 1/u * 1/2 * 1/(sqrt((2+x^2)/(2-x^2))) * (8x)/(2-x^2)^2`

`y^' = 1/sqrt((2+x^2)/(2-x)^2) * 1/2 * 1/(sqrt((2+x^2)/(2-x^2))) * (8x)/(2-x^2)^2`

`y^' = 1/2 * (1/sqrt((2+x^2)/(2-x^2)))^2 * (8x)/(2-x^2)^2`

`y^' = 1/2 * (2-x^2)/(2 + x^2) * (8x)/(2-x^2)^2`

This can be simplified to give

`y^' = 1/cancel(2) * color(red)(cancel(color(black)(2-x^2)))/(2 + x^2) * (cancel(8)^color(blue)(4)x)/(2-x^2)^color(red)(cancel(color(black)(2)))`

`y^' = color(green)((4x)/((2-x^2) * (2 + x^2)))`