Question

How do you differentiate `g(x) = sin (arctan (x/sqrt(3)))`?

Answer 1

`d/dx(sin(arctan(x/sqrt(3))))=3/(x^2+3)^(3/2)`

Explanation:

Let's first simplify `sin(arctan(x/sqrt(3)))`.

We know that `sin^2(theta)+cos^2(theta)=1`. Divide both sides by `sin(theta)^2` to get `1+1/tan^2(theta)=1/sin^2(theta)`, or `sin^2(theta)=1/(1+1/tan^2(theta))`.

Thus, `sin(theta)=sqrt(1/(1+1/tan^2(theta)))`.

So, using the above identity, `sin(arctan(x/sqrt(3)))=sqrt(1/(1+1/tan^2(arctan(x/sqrt(3)))))`.

Simplify:
`=sqrt(1/(1+1/(x^2/3)))=x/sqrt(x^2+3)`.

To differentiate this, use the quotient rule `d/dx(u/v)=((du)/dxv-(dv)/dxu)/v^2` to find the final answer

`d/dx(x/sqrt(x^2+3))=(sqrt(x^2+3)-x^2/sqrt(x^2+3))/(x^2+3)=3/(x^2+3)^(3/2)`