A calculus of the absurd

11.6 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 line7575 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).

11.6.1 Proof that the reciprocal function is decreasing

The reciprocal function7676 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.