10.7 Derivatives and slopes

I really wasn’t sure what to call this section other than \sayderivatives and slopes.

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

Something very similar is the case for derivatives. Firstly, for a straight line99 9 Read: \sayThe derivative of yy with respect to xx is equal to mm d⁢yd⁢x=m\frac{dy}{dx}=m.

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

10.7.1 Proof that the reciprocal function is decreasing

The reciprocal function1010 10 The function f⁢(x)=1xf(x)=\frac{1}{x}, can be differentiated,

dd⁢x⁢[1x]\displaystyle\frac{d}{dx}\left[\frac{1}{x}\right] =dd⁢x⁢[x−1]\displaystyle=\frac{d}{dx}\big{[}x^{-1}\big{]} (10.45)
=−x−2\displaystyle=-x^{-2} (10.46)
=−1x2\displaystyle=-\frac{1}{x^{2}} (10.47)

Because x2x^{2} is always greater than zero, it is also the case that

1x2>0\frac{1}{x^{2}}>0 (10.48)

Therefore, −1x2-\frac{1}{x^{2}} is always less than zero, so the reciprocal function is decreasing over its entire domain.