How is the derivative of $y = \frac{1}{x}$ - $\frac{1}{x^2}$?

Question

I'm trying to work out the derivatives of some simple functions via geometric models.

I pictured the function $y = \frac{1}{x}$ as a triangle with base $x$ and height $\frac{2}{x^2}$. The area of this triangle, therefore, is $\frac{1}{x}$.

Causing a differential $dx$ in the base $x$, we can see that the area of the triangle increases by $\frac{dx}{x^2}$. $$ d(\frac{1}{x}) = \frac{dx}{x^2} $$ That evaluates to: $$ \frac{d(\frac{1}{x})}{dx}=\frac{1}{x^2} $$

1. Why am I getting a $\frac{1}{x^2}$ here? Shouldn't it be $\frac{-1}{x^2}$?

Answer 1

I don't think you can use that model to derive the derivative of $1/x$ unless you already know the derivative of $1/x^2$. Which is a little silly - so note that the rest of this is not intended as a derivation of the derivative of $1/x$, just an explanation of what's going on with that model:

Informally, that $1/x^2$ is the change in the area due to the change in the base. When you change $x$ the area changes for two reasons: because the base changes and also because the height changes. It turns out that the change in the area due to the change in the height is $-2/x^2$, giving a total of $1/x^2-2/x^2=-1/x^2$.

Officially: You've written $$\frac1x=\frac12x\frac{2} {x^2}.$$So the product rule gives $$\left(\frac1x\right)'=\frac12x'\frac2{x^2}+\frac12x\left(\frac2{x^2}\right)'=\dots=\frac1{x^2}-\frac2{x^2}=\frac{-1}{x^2}.$$

Answer 2

Note that the area function is: $$y=\frac12\cdot x\cdot \frac{2}{x^2}.$$ When the base $x$ increases (decreases), the height decreases (increases) proportionally more, hence the function's area decreases (increases). You can work out the algebra to get the negative sign.

Answer 3

When you increase x, then (1/x) decreases, therefore "goes" in "opposite direction", hence minus sign.

In other words: if dx (change of x) is positive, then d(1/x) (changes of 1/x) must be negative, because what I wrote above.