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

2 Answers
Apr 1, 2016

Answer:

#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)#

Apr 1, 2016

Answer:

#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))#