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

2 Answers
Sasha P.
Sep 23, 2016

#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)#

sente
Oct 5, 2016

#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)#

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

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