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

2 Answers
Sep 23, 2016

1/(1+x^2) cos(arctanx)11+x2cos(arctanx)

Explanation:

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

=cos(arctanx)*(1/(1+x^2)) = 1/(1+x^2) cos(arctanx)=cos(arctanx)(11+x2)=11+x2cos(arctanx)

Oct 5, 2016

d/dx sin(arctan(x))=(x^2+1)^(-3/2)ddxsin(arctan(x))=(x2+1)32

Explanation:

Another method:

If we draw a right triangle with an angle thetaθ such that tan(theta) = xtan(θ)=x, then theta = arctan(x)θ=arctan(x). Using that triangle as a reference, we will find that sin(arctan(x)) = sin(theta) = x/sqrt(x^2+1)sin(arctan(x))=sin(θ)=xx2+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)


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