Question

What is the equation of the tangent line of `f(x)= x-x^-5` at `x=-4`?

Answer 1

`y = (1+5*(-4)^(-6))x - 6*(-4)^(-5)`

Explanation:

We have

`y = x - x^(-5)`

Derivating it we have

`dy/dx = 1 + 5x^(-6)`

So, we know the tangent line will have, for slope

`dy/dx|_(x=-4) = 1 + 5*(-4)^(-6)`

We also know, that, by definition, the tangent line will pass through the point `P(-4,f(-4))`, so we have

`(1+5*(-4)^(-6))*(-4) + b = -4 -(-4)^(-5)`
`b = -4 -(-4)^(-5) - (-4)(1+5\*(-4)^(-6))`
`b = (-4)(1 - (-4)^(-6) - 1 - 5\*(-4)^(-6))`
`b = (-4)(-6\*(-4)^(-6))`
`b = -6\*(-4)^(-5)`

So the tangent line is

`y = (1+5*(-4)^(-6))x - 6*(-4)^(-5)`