# A calculus of the absurd

#### 10.6 Derivatives and slopes

I really wasn’t sure what to call this section other than “derivatives and slopes”.

Here’s something you’ve probably noticed before - if we have a straight line in the form $$y = mx + c$$, then if $$m > 0$$ the line goes upwards (i.e. when $$x$$ increases, then $$y$$ increases too). If, however, $$m < 0$$ then the line goes downwards (i.e. when $$x$$ increases, $$y$$ becomes smaller)!

Something very similar is the case for derivatives. Firstly, for a straight line6969 Read: “The derivative of $$y$$ with respect to $$x$$ is equal to $$m$$” $$\frac {dy}{dx} = m$$.

It is the case that for every function $$f(x)$$ (which we can differentiate) that whenever $$\frac {dy}{dx} > 0$$ the function is increasing (i.e. as $$x$$ gets bigger, so does $$y$$). If $$\frac {dy}{dx} < 0$$, then the function is decreasing (that is, when $$x$$ gets bigger, $$y$$ gets smaller).

##### 10.6.1 Proof that the reciprocal function is decreasing

The reciprocal function7070 The function $$f(x) = \frac {1}{x}$$, can be differentiated,

\begin{align} \frac {d}{dx} \left [ \frac {1}{x} \right ] &= \frac {d}{dx} \big [ x^{-1} \big ] \\ &= - x^{-2} \\ &= - \frac {1}{x^2} \end{align}

Because $$x^2$$ is always greater than zero, it is also the case that

\begin{equation} \frac {1}{x^2} > 0 \end{equation}

Therefore, $$-\frac {1}{x^2}$$ is always less than zero, so the reciprocal function is decreasing over its entire domain.