Question

What's the derivative of `f(x) = sin (arctan (x/(sqrt(3))))`?

Answer 1

`cos(arctan(x/sqrt3))*1/sqrt3*1/(1+x^2/3)`

Explanation:

By the chain rule:

`y=sinu, u = arctan(x/sqrt3)`

`dy/(du)=cosu, (du)/dx=1/sqrt3*1/(1+x^2/3)`

`dy/dx = dy/(du)*(du)/dx`

`=cos(arctan(x/sqrt3))*1/sqrt3*1/(1+x^2/3)`

Answer 2

`f'(x) = 3/((x^2+3)sqrt(x^2+3))`

Explanation:

With `theta = arctan(x/sqrt3)` we get `sin theta = x/(sqrt(x^2+3)`.

I can't easily draw a picture of the triangle here, but there is an algebraic way to get this.
`tan theta = x/sqrt3` and `-pi/2 < theta < pi/2`
`rArr` `sec^2 theta = tan^2 theta +1 = x^2/3+1 = (x^2+3)/3`
`rArr` `cos^2 theta = 3/(x^2+3)`
`rArr` `sin^2 theta =1-cos^2 theta = 1-3/(x^2+3) = x^2/(x^2+3)`
`rArr` `sin theta = x/(sqrt(x^2+3)`.

So, we have

`f(x) = x/sqrt(x^2+3)`.

Use the quotient, power and chain rules to get

`f'(x) = 3/((x^2+3)sqrt(x^2+3))`