How do you find the derivative of #y = arctan(sqrt(3(x^2)-1))#?

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Answer 1 · 2017-06-18T10:11:30

#dy/dx=1/(xsqrt(3x^2-1))#

The first thing to know is the derivative of the function #y=tan^-1(u)#, where #u=g(x)# (think "the function on the inside").

#d/dx(tan^-1(u))=1/(1+u^2) (du)/dx#

#dy/dx=d/dx[tan^-1(sqrt(3x^2-1))]#

#=1/(1+(sqrt(3x^2-1))^2) d/dx[(3x^2-1)^(1/2)]#

#=1/(1+3x^2-1) (1/2(3x^2-1)^(-1/2)(6x))#

#=1/(cancel(3)x^cancel(2)) (cancel(3)cancel(x)(3x^2-1)^(-1/2))#

#=1/(xsqrt(3x^2-1))#