Question

How do you find the equation for the tangent line for `y=1/x` at `x=1`?

Answer 1

`y=1/x` can be expressed in an alternate form as
`y=x^(-1)`

The derivative of `y` gives the slope of the tangent line at a point `(x,y)`

`(dy)/(dx) = (-1)x^(-2)`
or (equivalently)
`(dy)/(dx) = - 1/x^2`

At x = 1
the slope is `(-1)/(1^2) = -1`

at `x=1`
`y = 1/x = 1`

The tangent has a slope of `(-1)` and passes through the point `(1,1)`

Using the slope-point form
`(y-1) = (-1)(x-1)`
`y-1 = 1-x`
`y=2-x`

The equation for the required tangent line is
`y=2-x`

Answer 2

You can alternatively use the Newton Approximation Method, which is:

`y = f(a) + f'(a)(x-a)`
where a is some arbitrary value and x is the typical unknown coordinate.

EX: If `f(x) = 1/x`:
`f(1) = 1/1` and `f'(1) = -1/x^2` at `x = 1`
`=> -1/1^2`.

So:
`y = 1/1 -(x-1)/1^2`

or

`y = 1 - (x - 1) = 2 - x` is your tangent line.

Power Rule: `(f'(1/x) = f'(x^-1) = -1*x^-2 = -1/x^2)`