How do you find the derivative of #Cos^-1 (3/x)#?

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Answer 1 · 2018-05-05T12:53:56

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

We have to know that,

#(arccos(x))' = -(1)/(sqrt(1-x^2))#

But in this case we have a chain rule to abide,

Where we an set #u = 3/x=3x^-1#

#(arccos(u))'=-(1)/(sqrt(1-u^2))*u'#

We now only need to find #u'#,

#u' = 3(-1*x^(-1-1))=-3x^-2=-3/x^2#

We will then have,

#(arccos(3/x))'=-(-3/x^2)/(sqrt(1-(3/x)^2)) = (3/x^2)/(sqrt(1-(3/x)^2))#