How do you find the equation of the tangent line to graph #y=x^(1/x)# at the point #(1, 1)#?

Question

3610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-05T19:29:26

The tangent line at #x=1# is #y=x#.

First, find the point on #y# that the tangent line will pass through by evaluating the function at #x=1#:

#y(1)=1^(1/1)=1#

The function passes through #(1,1)#.

To find the slope of the tangent line, differentiate #y#.

To do this, we will use logarithmic differentiation. That is, take the natural logarithm of both sides of the equation before taking the derivative:

#y=x^(1/x)#

#ln(y)=ln(x^(1/x))#

Rewriting using the log rule #log(A^B)=Blog(A)#:

#ln(y)=1/xln(x)#

#ln(y)=x^-1ln(x)#

Take the derivative of both sides of the equation. The left-hand side will require the chain rule. The right-hand side will use the product rule:

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

Since #y=x^(1/x)#:

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

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

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

Solving for the derivative:

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

The slope of the tangent line is:

#(dy/dx)_(x=1)=(1^(1/1)(1-ln(1)))/1^2=1(1-0)=1#

The line with slope #1# that passes through #(1,1)# can be given through the point-slope equation:

#y-1=1(x-1)#

Or:

#y=x#

Graphing #y=x# and #y=x^(1/x)#:

graph{(y-x^(1/x))(y-x)=0 [-1.628, 3.846, -0.475, 2.26]}

The line is tangent at #x=1#.