What is the equation of the line tangent to # f(x)=x^(1/x)# at # x=1#?

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Answer 1 · 2016-03-05T21:21:51

#y = x#

To find the tangent's slope just derivate the function.

We have
#y = x^(1/x)#

So take the log of both sides,
#ln(y) = ln(x)/x#

Derivate both sides,

#dy/dx*1/y= 1/x*d/dxln(x)+ln(x)*d/dx1/x#

#dy/dx*1/y = 1/x^2 - ln(x)*1/x^2#

#dy/dx = y*(1-ln(x))/x^2 = x^(1/x-2)*(1-ln(x))#

To find the slope at #x=1#, just switch it in, so:

#dy/dx = 1^(1/1-2)\*(1-ln(1)) = 1^(-1)*(1-0) = 1#

We know all lines follow the equation

#y = mx + b#

We know #y#, #x# and #m#, all that's left is discovering #b#

#(1)^(1/1) = 1*1 + b#
#1 = 1 + b#
#b = 0#

So the tangent line is

#y = x#

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