Ask a New Question
Menu

Determine the equation of the tangent at the point x=1 on the curve y= (x^x)^X

1 answer

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
Similar Questions
    1. answers icon 2 answers
  2. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  3. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
more similar questions