7.3 The derivative
Definition 7.16. The derivative of \(f\) at \(x\) is \[f'(x)=\frac {dy}{dx}=\lim _{h\to 0}\frac {f(x+h)-f(x)}{h},\] when the limit exists.
The fraction is the gradient of the line joining two points on the curve, at \(x\) and at \(x+h\). Letting \(h\to 0\) slides the second point onto the first, and the gradient of that chord becomes the gradient of the tangent. Note that setting \(h=0\) outright gives \(\frac {0}{0}\) — which is why a limit is needed rather than a substitution.
7.3.1 Differentiation from first principles
Solution. Step 1 — find \(f(x+h)\). Replace every \(x\) by \((x+h)\), keeping brackets: \[f(x+h)=(x+h)^2-3(x+h)=x^2+2xh+h^2-3x-3h.\]
Step 2 — subtract \(f(x)\). \[f(x+h)-f(x)=\left (x^2+2xh+h^2-3x-3h\right )-\left (x^2-3x\right )=2xh+h^2-3h.\] The terms without an \(h\) cancel, as they always must.
Step 3 — divide by \(h\). Every remaining term has a factor \(h\): \[\frac {f(x+h)-f(x)}{h}=\frac {h(2x+h-3)}{h}=2x+h-3.\]
Step 4 — let \(h\to 0\). \[f'(x)=\lim _{h\to 0}(2x+h-3)=2x-3.\]
Note 7.18. Step 3 is the point of the whole method. Until the \(h\) is cancelled, letting \(h\to 0\) gives \(\frac {0}{0}\); after cancelling, it is a simple substitution. If the \(h\) does not cancel, an algebra error has been made earlier.
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