Question

How do you differentiate `sin(arctan x)`?

Answer 1

`1/(1+x^2) cos(arctanx)`

Explanation:

`d/dx(sin(arctanx)) = cos (arctanx)d/dx(arctanx) =`

`=cos(arctanx)*(1/(1+x^2)) = 1/(1+x^2) cos(arctanx)`

Answer 2

`d/dx sin(arctan(x))=(x^2+1)^(-3/2)`

Explanation:

Another method:

If we draw a right triangle with an angle `theta` such that `tan(theta) = x`, then `theta = arctan(x)`. Using that triangle as a reference, we will find that `sin(arctan(x)) = sin(theta) = x/sqrt(x^2+1)`.

We can now differentiate using the quotient rule
`d/dx f(x)/g(x) = (f'(x)g(x) - f(x)g'(x))/[g(x)]^2`

`d/dx sin(arctan(x)) = d/dx x/sqrt(x^2+1)`

`= (sqrt(x^2+1) - x^2/sqrt(x^2+1))/(x^2+1)`

`=(x^2+1-x^2)/((x^2+1)sqrt(x^2+1))`

`(x^2+1)^(-3/2)`

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Note that this matches the other answer, as using the triangle, we can also see that `cos(arctan(x)) = 1/sqrt(x^2+1)`