Derivatives are not just a calculation drill: they are the tool that tells you where a function rises, where it falls, and how fast things change. In this chapter you will use \(f^{\prime}\) and \(f^{\prime\prime}\) to sketch curves, to find the best possible value in a real situation, to track quantities that change together, and to compute tricky limits.
1. Increasing and decreasing functions
Let \(f\) be differentiable on an interval \(I\). If \(f^{\prime}(x)>0\) for every \(x\) in \(I\), then \(f\) is increasing on \(I\). If \(f^{\prime}(x)<0\) for every \(x\) in \(I\), then \(f\) is decreasing on \(I\).
The derivative is the slope of the tangent line: a positive slope means the graph climbs from left to right, a negative slope means it descends. To find the intervals, solve \(f^{\prime}(x)=0\), place the solutions on a number line, and test one value in each piece.
Let \(f(x)=x^3-6x^2+9x+1\). Then \(f^{\prime}(x)=3x^2-12x+9=3(x-1)(x-3)\), which is zero at \(x=1\) and \(x=3\).
| Interval | \(x<1\) | \(1| \(x>3\) |
|
|---|---|---|---|
| \(f^{\prime}(x)=3(x-1)(x-3)\) | + | − | + |
| Behavior of \(f\) | increasing | decreasing | increasing |
So \(f\) increases on \((-\infty,1)\) and \((3,\infty)\) and decreases on \((1,3)\).
2. Critical points and extrema
A number \(c\) in the domain of \(f\) is a critical point if \(f^{\prime}(c)=0\) or \(f^{\prime}(c)\) does not exist. A local maximum (or minimum) at \(c\) means \(f(c)\) is the largest (or smallest) value of \(f\) near \(c\).
At a critical point \(c\): if \(f^{\prime}\) changes from \(+\) to \(-\), then \(f(c)\) is a local maximum; if it changes from \(-\) to \(+\), then \(f(c)\) is a local minimum; if the sign does not change, there is no extremum at \(c\).
In Example 1, \(f^{\prime}\) goes from \(+\) to \(-\) at \(x=1\), so \(f(1)=5\) is a local maximum; it goes from \(-\) to \(+\) at \(x=3\), so \(f(3)=1\) is a local minimum. On a closed interval \([a,b]\), the absolute extrema occur at critical points or at the endpoints, so compare all of those values.
\(f^{\prime}(c)=0\) does not guarantee an extremum. For \(f(x)=x^3\) we have \(f^{\prime}(0)=0\), yet the function keeps increasing through \(0\).
3. Concavity and inflection points
If \(f^{\prime\prime}(x)>0\) on an interval, the graph is concave up (it bends like a cup). If \(f^{\prime\prime}(x)<0\), it is concave down. An inflection point is a point of the graph where the concavity changes.
For Example 1, \(f^{\prime\prime}(x)=6x-12\), which is negative for \(x<2\) and positive for \(x>2\). The graph is concave down on \((-\infty,2)\), concave up on \((2,\infty)\), and \((2,\,f(2))=(2,3)\) is an inflection point. The second derivative test also helps: if \(f^{\prime}(c)=0\) and \(f^{\prime\prime}(c)>0\), then \(f(c)\) is a local minimum; if \(f^{\prime\prime}(c)<0\), a local maximum. Check: \(f^{\prime\prime}(1)=-6<0\) (maximum) and \(f^{\prime\prime}(3)=6>0\) (minimum).
4. Curve sketching
- Find the domain and the intercepts when they are easy.
- Compute \(f^{\prime}\), solve \(f^{\prime}=0\), and build the sign chart.
- Compute \(f^{\prime\prime}\) and find the concavity and inflection points.
- Evaluate \(f\) at every special point, then connect the points with a smooth curve that respects the signs.
The figure above is exactly this method applied to Example 1: a hill at \((1,5)\), a valley at \((3,1)\), and the bend at \((2,3)\), halfway between them.
5. The Mean Value Theorem
If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is at least one number \(c\) in \((a,b)\) such that \[ f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}. \]
In words: somewhere between \(a\) and \(b\), the instantaneous rate of change equals the average rate of change. Geometrically, some tangent line is parallel to the secant line.
For \(f(x)=x^2\) on \([1,4]\), the average rate is \(\dfrac{16-1}{4-1}=5\). Solve \(f^{\prime}(c)=2c=5\), so \(c=2.5\), which lies in \((1,4)\).
6. Optimization problems
- Draw a picture and name the variables.
- Write the quantity to optimize, then use the constraint to reduce it to one variable.
- State the allowed interval, find the critical points, and compare the values (including endpoints).
- Answer the question with units.
You have 200 ft of fencing for a rectangular garden that uses a long wall as one side, so only three sides need fencing. Let \(x\) be the length of each of the two sides perpendicular to the wall. The third side is \(200-2x\), so \(A(x)=x(200-2x)=200x-2x^2\) for \(0\le x\le100\).
\(A^{\prime}(x)=200-4x=0\) gives \(x=50\). Since \(A(0)=A(100)=0\) and \(A(50)=50\cdot100=5000\), the maximum area is \(5000\text{ ft}^2\) (about \(464.5\text{ m}^2\)), with dimensions \(50\text{ ft}\times100\text{ ft}\).
7. Related rates
When several quantities change with time and are linked by an equation, differentiate that equation with respect to \(t\) (using the chain rule), then substitute the known values after differentiating.
A 10 ft ladder leans against a wall. Its foot is \(x=6\) ft from the wall and slides away at \(\dfrac{dx}{dt}=2\) ft/s. How fast does the top move down?
The ladder, wall, and floor form a right triangle: \(x^2+y^2=100\). At that moment \(y=\sqrt{100-36}=8\). Differentiating: \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\), so \(\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}=-\dfrac{6}{8}\cdot2=-1.5\) ft/s. The top slides down at 1.5 ft/s (about 0.46 m/s).
Never plug the numbers in before you differentiate: a quantity that is changing must stay a variable until the derivative is taken!
8. L’Hôpital’s rule
Suppose \(\lim \dfrac{f(x)}{g(x)}\) has the form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\) (as \(x\to a\), or as \(x\to\pm\infty\)). If \(g^{\prime}(x)\ne0\) near \(a\) and \(\lim\dfrac{f^{\prime}(x)}{g^{\prime}(x)}\) exists, then \[ \lim\dfrac{f(x)}{g(x)}=\lim\dfrac{f^{\prime}(x)}{g^{\prime}(x)}. \]
\(\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x^2}\) is of the form \(\dfrac00\). Apply the rule: \(\dfrac{\sin x}{2x}\), still \(\dfrac00\); apply again: \(\dfrac{\cos x}{2}\to\dfrac12\). Also \(\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{x}=\lim\dfrac{1/x}{1}=0\).
Differentiate the numerator and the denominator separately. Do not use the quotient rule, and do not use L’Hôpital’s rule when the limit is not indeterminate.
9. Linear approximation
Near \(x=a\), a differentiable function is well approximated by its tangent line: \[ f(x)\approx L(x)=f(a)+f^{\prime}(a)(x-a). \]
Estimate \(\sqrt{26}\). Take \(f(x)=\sqrt x\) and \(a=25\): \(f(25)=5\) and \(f^{\prime}(25)=\dfrac{1}{2\sqrt{25}}=\dfrac1{10}\). So \(L(26)=5+\dfrac{1}{10}=5.1\). The calculator gives \(5.0990\), so the error is about \(0.001\).
Key takeaways
- \(f^{\prime}>0\) means increasing, \(f^{\prime}<0\) means decreasing; extrema occur at critical points where the sign of \(f^{\prime}\) changes.
- \(f^{\prime\prime}>0\) means concave up, \(f^{\prime\prime}<0\) means concave down; an inflection point is where concavity changes.
- MVT: some \(c\) has \(f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}\).
- Optimization: one variable, an interval, critical points, and endpoints.
- Related rates: differentiate with respect to \(t\) first, substitute after.
- L’Hôpital’s rule only for \(\dfrac00\) and \(\dfrac\infty\infty\); linearization: \(f(x)\approx f(a)+f^{\prime}(a)(x-a)\).
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