My Math Forum tangent to x^x

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 October 30th, 2008, 04:28 PM #1 Newbie   Joined: Oct 2008 Posts: 3 Thanks: 0 tangent to x^x the problem is find tangent line to x^x that passes through origo. i didnt get far: y=x^x lny=x*lnx y=e^(x*lnx) dy/dx = e^(x*lnx)*(x*1/x +lnx) = x^x(lnx +1) (is this derivative correct?) so, dy/dk must equal the slope of the line that passes though origo and so has equation y = (dy/dx)*x and im pretty much stuck here. Any help much appreciated!
 October 30th, 2008, 05:05 PM #2 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 Re: tangent to x^x Differentiate in the second line implicitly. y=x^x, so ln(y) = xln(x) <-----Here. Then separate variables and sub back for y. Too many late nights and I'm too tired to go on.
 November 2nd, 2008, 01:16 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,517 Thanks: 910 Math Focus: Elementary mathematics and beyond Re: tangent to x^x $y=x^x$ $ln(y)=xln(x)$ $\frac{y'}{y}=ln(x)+1$ $\mbox{Sub for y }=x^x$ $y'=x^xln(x)+x^x$ $\mbox{Equation of a line: }y= mx+b\mbox{ thus: }y=(x^xln(x)+x^x)x$ $\mbox{So: }y=x^{x+1}ln(x)+x^{x+1}$ $\mbox{Equation of curve: }y=x^x$ $\mbox{To find the point where the tangent line meets the curve: }x^x=x^{x+1}ln(x)+x^{x+1}$ $\mbox{So }x=1\mbox{ so the slope is 1, the y-intercept is 0}$ $\mbox{Equation of the line: }y=x$

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