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

Answer 1

Ezra Adams

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

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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Answer 2

Ezekiel Alexander

To find the derivative of the function ( \sec^{-1}(x^2 - x) ), you'll need to use the chain rule. The derivative of ( \sec^{-1}(u) ) with respect to ( x ) is ( \frac{1}{|u|\sqrt{u^2-1}} \frac{d}{dx}(u) ), where ( u = x^2 - x ). So, differentiate ( u = x^2 - x ) with respect to ( x ) to get ( \frac{d}{dx}(x^2 - x) ), and then apply the chain rule.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.