Question

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

Answer 1

`y=x`

Explanation:

We will need to differentiate the function to find the slope of the tangent line at the point `(1,1)`. To do so, take the natural logarithm of both sides.

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

This can be simplified using the logarithm rule: `log(a^b)=blog(a)`

`ln(y)=ln(x)/x`

Differentiate both sides. The left-hand side will use the chain rule, and the right-hand side will use the quotient rule.

`1/y(y^')=(1/x*x-ln(x)*1)/x^2`

`(y^')/y=(1-ln(x))/x^2`

`y^'=(y(1-ln(x)))/x^2`

`y^'=(x^(1/x)(1-ln(x)))/x^2`

We can simplify this using the rule `a^b/a^c=a^(b-c)`.

`y^'=x^(1/x-2)(1-ln(x))`

The slope of the line at `x=1` is:

`y^'(1)=1^(1/1-2)(1-ln(1))=1^-1(1-0)=1(1)=1`

The tangent line passes through `(1,1)` and has slope `1`. Thus its equation is

`y=1(x-1)+1" "=>" "y=x`

Graphed are the function and its tangent line at `(1,1)`:

graph{(y-x^(1/x))(y-x)=0 [-0.38, 10.72, -0.983, 4.567]}