How do you differentiate # g(x) = sin (arctan (x/sqrt(3))) #?

1 Answer
NickTheTurtle
Jan 9, 2018

#d/dx(sin(arctan(x/sqrt(3))))=3/(x^2+3)^(3/2)#

Explanation:

Let's first simplify #sin(arctan(x/sqrt(3)))#.

We know that #sin^2(theta)+cos^2(theta)=1#. Divide both sides by #sin(theta)^2# to get #1+1/tan^2(theta)=1/sin^2(theta)#, or #sin^2(theta)=1/(1+1/tan^2(theta))#.

Thus, #sin(theta)=sqrt(1/(1+1/tan^2(theta)))#.

So, using the above identity, #sin(arctan(x/sqrt(3)))=sqrt(1/(1+1/tan^2(arctan(x/sqrt(3)))))#.

Simplify:
#=sqrt(1/(1+1/(x^2/3)))=x/sqrt(x^2+3)#.

To differentiate this, use the quotient rule #d/dx(u/v)=((du)/dxv-(dv)/dxu)/v^2# to find the final answer

#d/dx(x/sqrt(x^2+3))=(sqrt(x^2+3)-x^2/sqrt(x^2+3))/(x^2+3)=3/(x^2+3)^(3/2)#

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