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What is the derivative of #y= arctan (sqrt 3x^2 -1)#?

1 Answer

Jun 27, 2018

#y'=(2sqrt3x)/(1+(sqrt3x^2+1)^2)#

Explanation:

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

Let #u=sqrt3x^2-1# , #(du)/dx=2sqrt3x#

#y'=(2sqrt3x)/(1+(sqrt3x^2+1)^2)#

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