Question

How do I find the derivative of `y = arctan(x/a) + ln sqrt((x-a)/(x+a))`?

Answer 1

`y^' = (2ax^2)/((a^2 + x^2)(x-a)(x+a))`

Explanation:

First, notice that you can simplify your starting function

`y = arctan(x/a) + ln(((x-a)/(x+a))^(1/2))`

`y = arctan(x/a) + 1/2 * ln((x-a)/(x+a))`

You will use implicit differentiation to find the derivative of `arctan(x/a)` first.

If you have `f = arctan(x/a)`, you can write

`tan(f) = x/a`

`sec^2(f) * (df)/dx = d/dx(x/a)`

`sec^2(f) * (df)/dx = 1/a`

This means that you have

`(df)/dx = 1/(a * sec^2(f))`

Use the trigonometric identity

`color(blue)(sec^2(x) = 1 + tan^2(x))`

to get

`(df)/dx = 1/(a * (1 + tan^2(f))) = 1/a * 1/(1 + (x/a)^2)`

`(df)/dx = 1/color(red)(cancel(color(black)(a))) * a^color(red)(cancel(color(black)(2)))/(a^2 + x^2) = a/(a^2 + x^2)`

Now use the chain rule for `lnu`, with `u = (x-a)/(x+a)` and the quotient rule to find

`d/dx(lnu) = d/(du)lnu * d/dx(u)`

`d/dx(lnu) = 1/u * d/dx((x-a)/(x+a))`

`d/dx(lnu) = 1/u * (([d/dx(x-a)] * (x+a) - (x-a) * d/dx(x+a))/(x+a)^2)`

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

`d/dx(ln((x-a)/(x+a))) = color(red)(cancel(color(black)((x+a))))/(x-a) * (color(red)(cancel(color(black)(x))) + a - color(red)(cancel(color(black)(a))) + a)/(x+a)^color(red)(cancel(color(black)(2)))`

`d/dx(ln((x-a)/(x+a)) = (2a)/((x-a)(x+a))`

So, your target derivative will thus be

`d/dx(y) = d/dx(arctan(x/a)) + 1/2 * d/dx(ln((x-a)/(x+a)))`

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

`y^' = [a(x-a)(x+a) + a(a^2 + x^2)]/((a^2 + x^2)(x-a)(x+a))`

`y^' = (ax^2 - color(red)(cancel(color(black)(a^3))) + ax^2 + color(red)(cancel(color(black)(a^3))))/((a^2 + x^2)(x-a)(x+a))`

Finally, you have

`y^' = color(green)((2ax^2)/((a^2 + x^2)(x-a)(x+a)))`