Differential Calculus
Rates of change, the derivative as an instantaneous rate, differentiation rules, the link between the sign of the derivative and monotonicity, and optimization.
- The slope between two points is \(\dfrac{\Delta y}{\Delta x}\).
- A tangent line touches a curve at a single nearby point.
6.1 Average rate of change
The average rate of change (a.r.c.) of \(f\) on \([a, b]\) is the slope of the secant line:
\[ \text{a.r.c.} = \frac{f(b) - f(a)}{b - a}. \]For \(f(x) = x^2\) on \([1, 3]\): \(\dfrac{f(3)-f(1)}{3-1} = \dfrac{9 - 1}{2} = 4.\)
6.2 Instantaneous rate of change (the derivative at a point)
The derivative of \(f\) at \(x = a\) is the limit of average rates as the interval shrinks:
\[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}. \]Geometrically, \(f'(a)\) is the slope of the tangent line at \(x=a\).
For \(f(x)=x^2\): \(\dfrac{(a+h)^2 - a^2}{h} = \dfrac{2ah + h^2}{h} = 2a + h \to 2a\). So \(f'(a) = 2a\).
6.3 The derivative function
Letting \(a\) vary gives the derivative function \(f'(x)\). The tangent line at \(x=a\) is
\[ y = f(a) + f'(a)(x - a). \]6.4 Rules of differentiation
Applying the power/quotient rule to \(f(x) = \dfrac{1}{x} = x^{-1}\) gives \(f'(x) = -x^{-2} = -\dfrac{1}{x^2}\).
\(f(x) = 3x^4 - 5x^2 + 7\Rightarrow f'(x) = 12x^3 - 10x.\)
6.5 Sign of the derivative and monotonicity
- \(f'(x) > 0\) on an interval \(\Rightarrow f\) is increasing there.
- \(f'(x) < 0\) on an interval \(\Rightarrow f\) is decreasing there.
- Where \(f'(x) = 0\) and the sign of \(f'\) changes, \(f\) has a local maximum (\(+\)→\(-\)) or minimum (\(-\)→\(+\)).
\(f(x) = x^3 - 3x\Rightarrow f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\). Zeros at \(x = \pm 1\): \(f\) increases on \(]{-\infty},-1[\), decreases on \(]-1,1[\), increases on \(]1,+\infty[\). Local max at \(x=-1\), local min at \(x=1\).
6.6 Optimization in a modeling context
Many applied problems ask for a maximum or minimum. The strategy:
- Write the quantity to optimize as a function of one variable (use constraints to eliminate others).
- Differentiate and solve \(f'(x) = 0\) for critical points.
- Use a sign chart (or context) to confirm max vs. min, and check endpoints if the domain is closed.
A rectangular pen against a wall uses \(20\ \text{m}\) of fencing on three sides. Maximize the area. Let width \(x\); then length \(= 20 - 2x\), so \(A(x) = x(20 - 2x) = 20x - 2x^2\). \(A'(x) = 20 - 4x = 0 \Rightarrow x = 5\). Since \(A'\) changes \(+\to-\), this is a maximum: \(A(5) = 5\cdot 10 = 50\ \text{m}^2\).
Practice problems
- Find the average rate of change of \(f(x)=x^3\) on \([0,2]\).
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\(\dfrac{8 - 0}{2} = 4\). - Differentiate \(g(x) = (2x+1)(x^2 - 3)\).
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Product rule: \(g'(x) = 2(x^2-3) + (2x+1)(2x) = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6\). - Find the equation of the tangent to \(f(x)=x^2\) at \(x=3\).
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\(f(3)=9,\ f'(x)=2x\Rightarrow f'(3)=6\). Tangent: \(y = 9 + 6(x-3) = 6x - 9\). - Find and classify the critical points of \(f(x)=x^2 - 4x + 1\).
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\(f'(x)=2x-4=0\Rightarrow x=2\); \(f'\) changes \(-\to+\), a minimum. \(f(2)=-3\).
Summary
- Average rate of change = secant slope; derivative = limit of a.r.c. = tangent slope.
- Power rule \((x^n)'=nx^{n-1}\), plus sum, product, and quotient rules.
- Sign of \(f'\) gives monotonicity; sign changes locate local extrema.
- Optimization: model, differentiate, solve \(f'=0\), then confirm.
- Differentiate \(f(x) = 5x^3 - 2x + 4\).
- On what interval is \(f(x) = x^2\) decreasing?