Calculus: The derivative of arctan(x/a)?

Question

Can you point me in the right direction? What is the derivative of arctan(x/a)? The whole multi-variable thing confuses me. Thanks.

Answer 1

I looked it up...been too long for me. :c)

Start with:

y = arctan(x/a)

That means:

tan y = x/a

Differentiating both sides implicitly (and knowing that (tan x)' = sec² x):

(sec² y) dy/dx = 1/a

We're looking for dy/dx, so:

dy/dx = 1/(a sec² y)

But we need to get rid of the y, so first use the fact that sec² = 1 + tan²:

dy/dx = 1/(a(1 + tan² y))

But, from above, we have that tan y = x/a, so substitute again:

dy/dx = 1/(a(1 + (x/a)²))

= 1/(a(1 + x²/a²))

= 1/(a((a² + x²)/a²))

= 1/((a² + x²)/a)

= a/(a² + x²)

--------------------------

An alternate way:

If you already knew that:

[tanˉ¹ x]' = 1/(1 + x²)

Then you could substitute u = x/a and use the chain rule:

[tanˉ¹ u]' = 1/(1 + u²) × u'

1/(1 + (x/a)²) × 1/a

and you can see that it will end up the same way.