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What is the derivative of this function #sec^-1(x^2-x)#?

1 Answer

Shwetank Mauria

Aug 5, 2016

#d/(dx)sec^(-1)(x^2-x)=(2x-1)/(|x^2-x|sqrt((x^2-x)^2-1))#

Explanation:

As derivative of #sec^(-1)x=1/(|x|sqrt(x^2-1))#, using chain rule,

#d/(dx)sec^(-1)(x^2-x)=1/(|x^2-x|sqrt((x^2-x)^2-1))xx(2x-1)#

= #(2x-1)/(|x^2-x|sqrt((x^2-x)^2-1))#

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