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

2 Answers
moutar
Apr 1, 2016

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)

Jim H
Apr 1, 2016

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))

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