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

Question

Christopher Allers · Accepted Answer

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#.

Evelyn Agard · Answer

undefined To find the equation of the tangent line to the graph of y=x^(1/x) at the point (1, 1), we can use the concept of differentiation.

First, we need to find the derivative of the function y=x^(1/x). Using the chain rule, the derivative is given by:

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

Simplifying this expression, we get:

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

Now, we substitute x=1 into the derivative to find the slope of the tangent line at the point (1, 1):

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

Since the slope of the tangent line is 0, the equation of the tangent line is simply y = 1.