Introduction to Calculus (Optional)

Exchanging \(a\) and \(b\) changes the sign of both \(f(b)-f(a)\) and \(b-a\), so it does not change the rate of change: it is enough to take \(a \lt b\). Then \(b - a \gt 0\), so the rate of change \(\dfrac{f(b)-f(a)}{b-a}\) has the sign of \(f(b)-f(a)\). Hence all the rates of change on \(I\) are positive or zero if and only if \(f(a) \leq f(b)\) whenever \(a \lt b\) in \(I\), that is, if and only if \(f\) is increasing on \(I\). The decreasing case is identical, with \(f(a) \geq f(b)\).

Constant, \(x\) and \(x^2\). For \(f(x) = k\), the rate of change is \(\dfrac{k - k}{h} = 0\); for \(f(x) = x\), it is \(\dfrac{(x+h) - x}{h} = 1\). For \(f(x) = x^2\): $$ \dfrac{(x+h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h \xrightarrow[h \to 0]{} 2x. $$ Inverse. For \(x \neq 0\) and \(h\) small enough that \(x + h \neq 0\), $$ \dfrac{\frac{1}{x+h} - \frac{1}{x}}{h} = \dfrac{\frac{x - (x+h)}{x(x+h)}}{h} = \dfrac{-h}{h\,x(x+h)} = -\dfrac{1}{x(x+h)} \xrightarrow[h \to 0]{} -\dfrac{1}{x \times x} = -\dfrac{1}{x^2}. $$ Square root. For \(x > 0\), we multiply the numerator and the denominator by \(\sqrt{x+h} + \sqrt{x}\) (the conjugate expression): $$ \dfrac{\sqrt{x+h}-\sqrt{x}}{h} = \dfrac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{h\left(\sqrt{x+h}+\sqrt{x}\right)} = \dfrac{(x+h) - x}{h\left(\sqrt{x+h}+\sqrt{x}\right)} = \dfrac{1}{\sqrt{x+h}+\sqrt{x}} \xrightarrow[h \to 0]{} \dfrac{1}{2\sqrt{x}}. $$ Powers. For \(f(x) = x^3\), since \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\), $$ \dfrac{(x+h)^3 - x^3}{h} = \dfrac{h(3x^2 + 3xh + h^2)}{h} = 3x^2 + 3xh + h^2 \xrightarrow[h \to 0]{} 3x^2. $$ The general formula \((x^n)' = nx^{n-1}\) for \(n \in \mathbb{Z}\) is admitted: for \(n \geq 1\) the computation follows the same pattern, and for \(n \leq -1\) it follows from the case \(n \geq 1\) and the derivative of the reciprocal of a function.

Proof for if \(f(x)=u(x)+v(x)\), then \(f'(x)=u'(x)+v'(x)\):
We evaluate the limit of the rate of change.$$\begin{aligned}\dfrac{f(x+h)-f(x)}{h} &= \dfrac{\left[u(x+h)+v(x+h)\right] - \left[u(x)+v(x)\right]}{h} \\ &= \dfrac{\left[u(x+h)-u(x)\right] + \left[v(x+h)-v(x)\right]}{h} \\ &= \dfrac{u(x+h)-u(x)}{h} + \dfrac{v(x+h)-v(x)}{h} \\ &\xrightarrow[h \to 0]{ } u'(x) + v'(x)\quad \text{(sum of limits)}.\end{aligned}$$ So \( f'(x) = u'(x)+v'(x) \).
The other proofs are done in the exercises.

Let \(B(x,y)\) be any point on the tangent line. The point of tangency is \(A(a,f(a))\), which also belongs to the tangent line.
The gradient (also called slope) of a line is given by the formula \(m = \dfrac{\Delta y}{\Delta x}\). Using points \(A\) and \(B\), the slope of the tangent line is:$$ m = \dfrac{y-f(a)}{x-a}\quad \text{for }x\neq a. $$ By definition, the slope of the tangent to the curve \(y=f(x)\) at the point \(x=a\) is the value of the derivative at that point, so:$$ m = f'(a). $$ Equating the two expressions for the slope, we get:$$ f'(a) = \dfrac{y-f(a)}{x-a}. $$ Multiplying both sides by \((x-a)\) gives the point-gradient form (also called point-slope form) of the equation:$$ y - f(a) = f'(a)(x-a). $$ Making \(y\) the subject gives:$$ y = f'(a)(x-a) + f(a). $$

The tangent has equation \(y = f'(a)(x - a) + f(a)\). A point of this line lies on the \(x\)-axis when \(y = 0\): \begin{align*} f'(a)(x - a) + f(a) = 0 &\iff f'(a)(x - a) = -f(a)
&\iff x - a = -\dfrac{f(a)}{f'(a)} && \text{(division by } f'(a) \neq 0\text{)}
&\iff x = a - \dfrac{f(a)}{f'(a)}. \end{align*} When \(f'(a) = 0\), the division cannot be made: the tangent is horizontal (it is the \(x\)-axis itself when \(f(a) = 0\)).

By definition of the derivative at \(a\), the rate of change \(\dfrac{f(a+h) - f(a)}{h}\) tends to \(f'(a)\) as \(h\) tends to \(0\). Set, for \(h \neq 0\), $$\varepsilon(h) = \dfrac{f(a+h) - f(a)}{h} - f'(a).$$ Then \(\varepsilon(h)\) tends to \(0\), and multiplying by \(h\) gives \(f(a+h) - f(a) - f'(a)\,h = h\,\varepsilon(h)\), that is \(f(a+h) = f(a) + f'(a)\,h + h\,\varepsilon(h)\). The error \(h\,\varepsilon(h)\) is the product of two quantities that both tend to \(0\): it is negligible compared with \(h\).

Let \(a \in I\) and suppose that \(f\) is increasing on \(I\). For \(x \in I\) with \(x \neq a\), consider the rate of change \(\dfrac{f(x)-f(a)}{x-a}\).

• If \(x < a\): since \(f\) is increasing, \(f(x) \leq f(a)\), so \(f(x) - f(a) \leq 0\) and \(x - a < 0\). The quotient of two numbers of the same sign is positive or zero.
• If \(x > a\): \(f(x) \geq f(a)\), so \(f(x) - f(a) \geq 0\) and \(x - a > 0\). Again the quotient is positive or zero.

So the rate of change is greater than or equal to \(0\) for every \(x \neq a\). We accept without proof that this inequality is preserved when \(x\) approaches \(a\): its limit \(f'(a)\) satisfies \(f'(a) \geq 0\). As \(a\) is any point of \(I\), we get \(f'(x) \geq 0\) for all \(x \in I\).
If \(f\) is decreasing, the same reasoning gives a quotient less than or equal to \(0\), hence \(f'(a) \leq 0\). If \(f\) is constant, \(f(x) - f(a) = 0\): the rate of change is \(0\) and so is \(f'(a)\).

\(f'(x) = 2ax + b\) is a linear function with gradient \(2a\); it vanishes at \(x_0 = -\dfrac{b}{2a}\). If \(a > 0\), then \(2a > 0\), so \(f'(x) < 0\) for \(x < x_0\) and \(f'(x) > 0\) for \(x > x_0\): \(f\) is decreasing then increasing, and \(f'\) vanishes at \(x_0\) while changing sign from \(-\) to \(+\), so \(f\) has a minimum at \(x_0\). If \(a < 0\), all the signs are reversed: \(f\) is increasing then decreasing, with a maximum at \(x_0\).