Asked by Lisa
Determine the equation of the tangent at the point x=1 on the curve y= (x^x)^X
Answers
Answered by
oobleck
y = (x^x)^x = x^(x^2)
y(1) = 1
y' = x^(x^2) (x + 2x lnx)
y'(1) = 1(1+2*0) = 2
so the line is
y-1 = 1(x-1)
or, y=x
confirm at
https://www.wolframalpha.com/input/?i=plot+y%3D%28x%5Ex%29%5Ex%2C+y%3Dx
y(1) = 1
y' = x^(x^2) (x + 2x lnx)
y'(1) = 1(1+2*0) = 2
so the line is
y-1 = 1(x-1)
or, y=x
confirm at
https://www.wolframalpha.com/input/?i=plot+y%3D%28x%5Ex%29%5Ex%2C+y%3Dx
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