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

1 Answer
Stefan V.
Aug 25, 2015

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)))

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