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

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 $\frac{-1}{x^2}$?
• You appear to be working as if the height remained constant, while it is clearly dependent on the length of the basis.
– user228113
Commented May 7, 2018 at 10:40

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 $$\left(\frac1x\right)'=\frac12x'\frac2{x^2}+\frac12x\left(\frac2{x^2}\right)'=\dots=\frac1{x^2}-\frac2{x^2}=\frac{-1}{x^2}.$$

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.

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.