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").

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

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

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

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

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

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

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