Chapter 2: Differentiation Rules
In Chapter 1 we studied limits and continuity, the foundational ideas that make calculus possible. Now we turn to the first of the two central operations of calculus: differentiation. The derivative measures how a function changes as its input changes, and it has applications throughout science, engineering, economics, and beyond. In this chapter we define the derivative precisely using limits, develop the rules that make computing derivatives efficient, and apply those rules to polynomial, trigonometric, exponential, and logarithmic functions.
🎯 Learning Objectives
After completing this chapter, you will be able to:
- Compute derivatives using the limit definition (difference quotient)
- Apply the power, constant, sum, and difference rules to differentiate polynomials and radical expressions
- Differentiate trigonometric, exponential, and logarithmic functions
- Apply the product and quotient rules to differentiate products and quotients of functions
- Find higher-order derivatives and interpret velocity and acceleration physically
- Identify points where a function fails to be differentiable and explain why
- Find equations of tangent and normal lines to curves at given points
2.1 The Derivative as a Limit
Consider a function $y = f(x)$ and two points on its graph: $(x, f(x))$ and $(x+h, f(x+h))$. The slope of the straight line connecting these two points is the difference quotient:
$$\text{slope of secant} = \frac{f(x+h) - f(x)}{h}$$This secant line gives an average rate of change of $f$ over the interval from $x$ to $x+h$. As we let $h$ shrink toward zero, the secant line rotates and approaches the tangent line to the curve at the point $(x, f(x))$. The slope of that tangent line is the instantaneous rate of change of $f$ at $x$, and this is precisely what we call the derivative.
Average Rate of Change vs. Instantaneous Rate of Change
These two ideas — slope between two points and slope at one point — are the heart of differential calculus. The table below shows how they relate and what each one is called.
| Average Rate of Change | Instantaneous Rate of Change | |
|---|---|---|
| Also called | Mean value · Average slope · Difference quotient | Derivative · Differential coefficient · $f'(x)$ |
| Geometric meaning | Slope of the secant line through two points on the curve | Slope of the tangent line at a single point on the curve |
| Formula | $\dfrac{f(b)-f(a)}{b-a} = \dfrac{\Delta y}{\Delta x}$ | $\displaystyle\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h} = f'(x)$ |
| Notation | $\dfrac{\Delta y}{\Delta x}$ | $\dfrac{dy}{dx}$, $f'(x)$, $D_x f$ |
| Requires | Two distinct points: $(a,\, f(a))$ and $(b,\, f(b))$ | One point: $(x,\, f(x))$, obtained by taking a limit |
| Key idea | How much did $f$ change on average over an interval? | How fast is $f$ changing right now at a single input? |
Figure 2.1a — Average Rate of Change. The blue secant line passes through two points $A$ and $B$. Its slope $= \dfrac{\Delta y}{\Delta x}$ is the average rate of change. Drag the slider to move $B$ closer to $A$.
Figure 2.1b — Instantaneous Rate of Change. As $B \to A$ (i.e. $h \to 0$), the secant line becomes the red tangent line. Its slope is $f'(a)$ — the derivative.
AP Exam Tip: Average rate of change problems give you two points — just compute $\dfrac{f(b)-f(a)}{b-a}$. Instantaneous rate of change problems ask for a derivative at a point — find $f'(x)$ then evaluate. The AP exam frequently tests whether you know which one is being asked for.
Definition 2.1 — The Derivative (Three Equivalent Forms)
The derivative of a function $f$ is defined by any of the following equivalent limit expressions. All three describe the same instantaneous rate of change.
Form 1 — The $h$-form (derivative as a function of $x$):
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$Form 2 — The $\Delta x$-form (same idea, alternate notation):
$$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$Form 3 — The point form (derivative at a specific point $x = a$):
$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$The limit must exist for $f$ to be differentiable at that point. Forms 1 and 2 are interchangeable ($\Delta x$ plays the role of $h$). Form 3 is especially useful when evaluating $f'$ at a specific value $a$ — you see it on AP free-response problems where a table of values is given.
Equivalent notations for the derivative include $\dfrac{dy}{dx}$, $\dfrac{df}{dx}$, and $D_x[f(x)]$. Each emphasizes a slightly different perspective: Leibniz notation $\dfrac{dy}{dx}$ highlights the ratio-of-changes origin, while prime notation $f'(x)$ is compact and convenient for algebraic work.
Geometric Interpretation
Geometrically, $f'(a)$ is the slope of the line tangent to $y = f(x)$ at $x = a$. The equation of this tangent line is:
$$y - f(a) = f'(a)(x - a)$$If $f'(a) > 0$, the function is increasing at $x = a$. If $f'(a) < 0$, the function is decreasing. If $f'(a) = 0$, the tangent line is horizontal, which often corresponds to a local maximum, local minimum, or inflection point.
Figure 2.1 — Tangent line via secant limit. Drag the slider to move the point of tangency along $f(x) = x^3 - 3x$. The dashed red line is the tangent; the dashed green curve is the derivative $f'(x)$. The slope of the tangent at any point equals the value of $f'$ there.
Computing Derivatives from the Definition
Before we develop shortcut rules, it is important to be able to compute derivatives directly from the limit definition. The standard four-step procedure is:
- Write out $f(x+h)$ by substituting $(x+h)$ everywhere $x$ appears.
- Compute the difference $f(x+h) - f(x)$ and simplify.
- Divide by $h$ and cancel or simplify so that direct substitution of $h = 0$ is possible.
- Evaluate the limit as $h \to 0$.
Worked Example 2.1 — Derivative of a Polynomial from the Definition
Find $f'(x)$ for $f(x) = 3x^2 - 5x + 1$ using the limit definition.
Solution. We compute:
$$f(x+h) = 3(x+h)^2 - 5(x+h) + 1 = 3x^2 + 6xh + 3h^2 - 5x - 5h + 1$$
The difference is:
$$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h + 1) - (3x^2 - 5x + 1) = 6xh + 3h^2 - 5h$$
Dividing by $h$:
$$\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h - 5)}{h} = 6x + 3h - 5$$
Taking the limit:
$$f'(x) = \lim_{h \to 0}(6x + 3h - 5) = \boxed{6x - 5}$$
Try It 2.1 — Use the limit definition to find $f'(x)$ for $f(x) = 2x^2 - 4x$.
Show Hint
Follow the same four steps. Expand $f(x+h) = 2(x+h)^2 - 4(x+h)$, form the difference $f(x+h) - f(x)$, factor out $h$, cancel, and take the limit. Answer: $f'(x) = 4x - 4$.
Worked Example 2.2 — Derivative of a Square Root Function
Find $f'(x)$ for $f(x) = \sqrt{x}$ using the limit definition.
Solution. We write:
$$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$
Direct substitution gives $\frac{0}{0}$, so we rationalize the numerator by multiplying by the conjugate:
$$= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$
$$= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$
Now substituting $h = 0$:
$$f'(x) = \frac{1}{2\sqrt{x}}, \quad x > 0$$
This result is consistent with the power rule applied to $x^{1/2}$, which we develop in the next section.
Try It 2.2 — Use the limit definition to find $f'(x)$ for $f(x) = \dfrac{1}{x}$.
Show Hint
Write $\dfrac{f(x+h)-f(x)}{h} = \dfrac{\frac{1}{x+h} - \frac{1}{x}}{h}$. Combine the fractions in the numerator over a common denominator $x(x+h)$, simplify, cancel $h$, and take the limit. Answer: $f'(x) = -\dfrac{1}{x^2}$.
When Is a Function Not Differentiable?
Not every continuous function has a derivative everywhere. There are three classic types of points where a function is continuous but fails to be differentiable, plus the trivial case of a discontinuity.
Definition 2.2 — Types of Non-Differentiable Points
A function $f$ is not differentiable at $x = a$ if the limit $\displaystyle\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ does not exist. This occurs at:
- Corners: the left-hand and right-hand limits of the difference quotient exist but are unequal (e.g., $f(x) = |x|$ at $x = 0$).
- Cusps: the difference quotient tends to $+\infty$ from one side and $-\infty$ from the other (e.g., $f(x) = x^{2/3}$ at $x = 0$).
- Vertical tangents: the difference quotient tends to $\pm\infty$ from both sides (e.g., $f(x) = x^{1/3}$ at $x = 0$).
- Discontinuities: if $f$ is not continuous at $a$, it cannot be differentiable there either.
Figure 2.2a — Corner. $f(x)=|x|$ is continuous everywhere but has no tangent at the origin.
Figure 2.2b — Cusp. $f(x)=x^{2/3}$ has a sharp point at the origin where the slope is infinite.
Figure 2.2c — Vertical tangent. $f(x)=x^{1/3}$ has a vertical tangent at the origin; the slope is undefined.
Theorem 2.1 — Differentiability Implies Continuity
If $f$ is differentiable at $x = a$, then $f$ is continuous at $x = a$.
Proof sketch. We want to show $\lim_{x \to a} f(x) = f(a)$, i.e., $\lim_{h \to 0} [f(a+h) - f(a)] = 0$. Write:
$$f(a+h) - f(a) = \frac{f(a+h) - f(a)}{h} \cdot h$$
Taking the limit as $h \to 0$: $\lim_{h \to 0}[f(a+h) - f(a)] = f'(a) \cdot 0 = 0$. Therefore $\lim_{h \to 0} f(a+h) = f(a)$. $\square$
Important: The Converse Is False
A function can be continuous at a point without being differentiable there. The function $f(x) = |x|$ is continuous at $x = 0$ (it has no gaps or jumps), but it has a corner at the origin, so $f'(0)$ does not exist. Continuity is necessary but not sufficient for differentiability.
AP Exam Tip: Differentiability implies continuity, but NOT vice versa. $|x|$ is the classic counterexample: continuous at $x = 0$ but not differentiable there because of the corner. On the exam you may be asked to identify non-differentiable points on a graph — look for corners, cusps, vertical tangents, and discontinuities.
2.2 Basic Differentiation Rules
Using the limit definition for every derivative would be tedious. Fortunately, a small collection of rules allows us to differentiate most functions quickly. Each of these rules can be proved rigorously using the limit definition, but we state them as theorems and focus on their application.
Theorem 2.2 — Constant Rule
If $c$ is a constant, then $\dfrac{d}{dx}[c] = 0$.
This makes intuitive sense: a constant function has a horizontal graph, so its slope is zero everywhere.
Theorem 2.3 — Power Rule
If $n$ is any real number, then
$$\frac{d}{dx}[x^n] = nx^{n-1}$$
The power rule works not just for positive integers but for all real exponents: negative integers, fractions, and irrational numbers. For instance, $\dfrac{d}{dx}[x^{-3}] = -3x^{-4}$ and $\dfrac{d}{dx}[x^{\pi}] = \pi x^{\pi - 1}$.
Figure 2.3 — The Power Rule. Adjust the exponent $n$ with the slider. The blue curve is $f(x) = x^n$ and the dashed red curve is its derivative $f'(x) = nx^{n-1}$. Notice how the derivative of a degree-$n$ curve is a degree-$(n-1)$ curve.
Theorem 2.4 — Constant Multiple Rule
If $c$ is a constant and $f$ is differentiable, then
$$\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$$
Constants factor out of the derivative. This follows immediately from the limit definition, since we can factor a constant out of a limit.
Theorem 2.5 — Sum and Difference Rules
If $f$ and $g$ are both differentiable, then
$$\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$$
The derivative of a sum (or difference) is the sum (or difference) of the derivatives. Combined with the power rule and constant multiple rule, this lets us differentiate any polynomial term by term.
Worked Example 2.3 — Differentiating a Polynomial
Find $\dfrac{dy}{dx}$ for $y = 4x^5 - 7x^3 + 2x - 9$.
Solution. Applying the power rule term by term:
$$\frac{dy}{dx} = 4 \cdot 5x^4 - 7 \cdot 3x^2 + 2 \cdot 1 - 0 = \boxed{20x^4 - 21x^2 + 2}$$
Try It 2.3 — Differentiate $y = 3x^4 - 5x^2 + 7x - 2$.
Show Hint
Apply the power rule to each term independently: $\dfrac{dy}{dx} = 12x^3 - 10x + 7$.
Worked Example 2.4 — Rewriting Before Differentiating
Find $f'(x)$ for $f(x) = \dfrac{3}{x^2} - \dfrac{5}{\sqrt{x}} + 8$.
Solution. First rewrite using negative and fractional exponents:
$$f(x) = 3x^{-2} - 5x^{-1/2} + 8$$
Now apply the power rule to each term:
$$f'(x) = 3(-2)x^{-3} - 5\!\left(-\tfrac{1}{2}\right)x^{-3/2} + 0 = -6x^{-3} + \frac{5}{2}x^{-3/2}$$
Rewriting in radical/fraction form:
$$f'(x) = -\frac{6}{x^3} + \frac{5}{2x\sqrt{x}}$$
Try It 2.4 — Differentiate $f(x) = \dfrac{4}{x^3} + 2\sqrt{x} - 1$.
Show Hint
Rewrite as $f(x) = 4x^{-3} + 2x^{1/2} - 1$. Then apply the power rule: $f'(x) = -12x^{-4} + x^{-1/2} = -\dfrac{12}{x^4} + \dfrac{1}{\sqrt{x}}$.
Worked Example 2.5 — Finding a Tangent Line
Find the equation of the tangent line to $y = x^3 - 4x + 2$ at the point where $x = 1$.
Solution. First, find the $y$-coordinate: $y(1) = 1 - 4 + 2 = -1$. So the point is $(1, -1)$.
Next, compute the derivative: $y' = 3x^2 - 4$. At $x = 1$:
$$y'(1) = 3(1)^2 - 4 = -1$$
Using point-slope form:
$$y - (-1) = -1(x - 1) \implies \boxed{y = -x}$$
Try It 2.5 — Find the equation of the tangent line to $y = 2x^2 - 3x + 1$ at $x = 2$.
Show Hint
Compute $y(2) = 8 - 6 + 1 = 3$. Then $y' = 4x - 3$, so $y'(2) = 5$. Using point-slope: $y - 3 = 5(x - 2)$, which gives $y = 5x - 7$.
AP Exam Tip: Always simplify expressions to power form ($x^n$) before differentiating. The power rule does not work directly on radical or fraction form. Rewrite $\dfrac{1}{x^3}$ as $x^{-3}$ and $\sqrt[3]{x^2}$ as $x^{2/3}$ first, then differentiate.
2.3 Derivatives of Trigonometric Functions
The six trigonometric functions arise constantly in calculus, so knowing their derivatives is essential. We start by deriving the most fundamental one from the limit definition, then list all six.
Derivation of $\dfrac{d}{dx}[\sin x]$
Using the limit definition:
$$\frac{d}{dx}[\sin x] = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$$
Apply the angle addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$:
$$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$
$$= \lim_{h \to 0} \left[\sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}\right]$$
We use two special limits established in Chapter 1:
$$\lim_{h \to 0} \frac{\sin h}{h} = 1, \qquad \lim_{h \to 0} \frac{\cos h - 1}{h} = 0$$
Therefore:
$$\frac{d}{dx}[\sin x] = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$
All Six Trigonometric Derivatives
A similar analysis for the remaining five functions yields the complete list:
| $f(x)$ | $f'(x)$ |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ |
Figure 2.4 — $\sin x$ and its derivative $\cos x$. The blue curve is $y = \sin x$ and the dashed red curve is $y = \cos x = \dfrac{d}{dx}[\sin x]$. At each $x$, the slope of sin $x$ equals the height of cos $x$. Observe that $\cos x = 0$ exactly where $\sin x$ has a horizontal tangent (peaks and troughs).
AP Exam Tip: The derivatives of the "co-" functions (cosine, cotangent, cosecant) always carry a negative sign. This is the single most reliable pattern for memorizing the trig derivative table. On a free-response question, a missing negative sign is a very common error.
Worked Example 2.6 — Combining Trig and Power Rules
Differentiate $g(x) = 5\sin x - 2\cos x + 3x^2$.
Solution. Using the sum rule, constant multiple rule, and trig derivatives:
$$g'(x) = 5\cos x - 2(-\sin x) + 6x = \boxed{5\cos x + 2\sin x + 6x}$$
Try It 2.6 — Differentiate $h(x) = 4\cos x + 3\sin x - x^3$.
Show Hint
Each term differentiates independently. $h'(x) = -4\sin x + 3\cos x - 3x^2$.
Worked Example 2.7 — Tangent Line to a Trig Function
Find the slope of the tangent line to $y = \tan x$ at $x = \dfrac{\pi}{4}$.
Solution. Since $\dfrac{d}{dx}[\tan x] = \sec^2 x$:
$$y'\!\left(\frac{\pi}{4}\right) = \sec^2\!\left(\frac{\pi}{4}\right) = \left(\frac{1}{\cos(\pi/4)}\right)^2 = \left(\frac{1}{\sqrt{2}/2}\right)^2 = (\sqrt{2})^2 = \boxed{2}$$
Try It 2.7 — Find the slope of the tangent line to $y = \sec x$ at $x = \dfrac{\pi}{3}$.
Show Hint
$\dfrac{d}{dx}[\sec x] = \sec x \tan x$. At $x = \dfrac{\pi}{3}$: $\sec\dfrac{\pi}{3} = 2$ and $\tan\dfrac{\pi}{3} = \sqrt{3}$. So the slope is $2\sqrt{3}$.
2.4 Derivatives of Exponential and Logarithmic Functions
Exponential and logarithmic functions play a central role in modeling growth and decay. Their derivative formulas are among the most elegant in all of calculus.
Theorem 2.6 — Derivative of the Natural Exponential
$$\frac{d}{dx}[e^x] = e^x$$
The function $e^x$ is its own derivative. This remarkable property is, in fact, one way to define the number $e \approx 2.71828$: it is the unique base $a$ for which $\lim_{h \to 0} \dfrac{a^h - 1}{h} = 1$, which forces $\dfrac{d}{dx}[a^x]\big|_{x=0} = 1$.
Figure 2.5 — $e^x$ is its own derivative. For $f(x) = e^x$, the slope of the tangent line at any point equals the $y$-coordinate. Move the slider to verify this remarkable property at different points along the curve.
AP Exam Tip: Do NOT apply the power rule to $e^x$. The power rule $\dfrac{d}{dx}[x^n] = nx^{n-1}$ applies to a variable base with constant exponent (like $x^3$), not a constant base with a variable exponent (like $e^x$). Writing $\dfrac{d}{dx}[e^x] = xe^{x-1}$ is a very common and costly error.
Theorem 2.7 — Derivative of the Natural Logarithm
$$\frac{d}{dx}[\ln x] = \frac{1}{x}, \quad x > 0$$
Proof via implicit differentiation. If $y = \ln x$, then $e^y = x$. Differentiating both sides with respect to $x$ gives $e^y \dfrac{dy}{dx} = 1$, so $\dfrac{dy}{dx} = \dfrac{1}{e^y} = \dfrac{1}{x}$. $\square$
Theorem 2.8 — General Exponential and Logarithmic Derivatives
For any constant $a > 0$, $a \neq 1$:
$$\frac{d}{dx}[a^x] = a^x \ln a, \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$$
The general exponential formula follows by writing $a^x = e^{x \ln a}$ and applying the chain rule (Chapter 3). The logarithmic formula follows from the change-of-base identity $\log_a x = \dfrac{\ln x}{\ln a}$.
Worked Example 2.8 — Differentiating Exponential Expressions
Find $\dfrac{dy}{dx}$ for $y = 3e^x + 7 \cdot 2^x - 4x^3$.
Solution. Applying the rules term by term:
$$\frac{dy}{dx} = 3e^x + 7 \cdot 2^x \ln 2 - 12x^2$$
Try It 2.8 — Find $\dfrac{dy}{dx}$ for $y = 5e^x - 3 \cdot 3^x + x^2$.
Show Hint
The general exponential rule gives $\dfrac{d}{dx}[3^x] = 3^x \ln 3$. So $\dfrac{dy}{dx} = 5e^x - 3 \cdot 3^x \ln 3 + 2x$.
Worked Example 2.9 — Differentiating Logarithmic Expressions
Find $f'(x)$ for $f(x) = 4\ln x - \log_3 x + x^2$.
Solution. Applying Theorems 2.7 and 2.8:
$$f'(x) = \frac{4}{x} - \frac{1}{x \ln 3} + 2x$$
We can combine the first two terms over a common denominator if desired:
$$f'(x) = \frac{4\ln 3 - 1}{x \ln 3} + 2x$$
Try It 2.9 — Find $f'(x)$ for $f(x) = 2\ln x + \log_2 x - x$.
Show Hint
Apply Theorem 2.8 to the $\log_2$ term with $a = 2$: $\dfrac{d}{dx}[\log_2 x] = \dfrac{1}{x \ln 2}$. So $f'(x) = \dfrac{2}{x} + \dfrac{1}{x \ln 2} - 1$.
2.5 Product and Quotient Rules
When a function is formed as a product or quotient of two simpler functions, we need dedicated rules. A common misconception is that the derivative of a product is the product of the derivatives — this is false. The correct rules are given below.
Theorem 2.9 — Product Rule
If $f$ and $g$ are both differentiable, then
$$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$$
Proof sketch. Start with the limit definition and use the algebraic trick of adding and subtracting $f(x+h)g(x)$ in the numerator:
$$\frac{d}{dx}[fg] = \lim_{h \to 0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h}$$
$$= \lim_{h \to 0}\left[f(x+h)\cdot\frac{g(x+h)-g(x)}{h} + g(x)\cdot\frac{f(x+h)-f(x)}{h}\right]$$
Since $f$ is differentiable (hence continuous), $\lim_{h\to 0} f(x+h) = f(x)$. The two difference quotients tend to $g'(x)$ and $f'(x)$ respectively, giving the result. $\square$
Theorem 2.10 — Quotient Rule
If $f$ and $g$ are both differentiable and $g(x) \neq 0$, then
$$\frac{d}{dx}\!\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$
Proof sketch. Starting from the limit definition and combining fractions over a common denominator:
$$\frac{d}{dx}\!\left[\frac{f}{g}\right] = \lim_{h \to 0}\frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} = \lim_{h \to 0}\frac{f(x+h)g(x) - f(x)g(x+h)}{h \cdot g(x+h) \cdot g(x)}$$
Adding and subtracting $f(x)g(x)$ in the numerator and using similar limit arguments yields the formula. $\square$
AP Exam Tip: "Low d-high minus high d-low, square the bottom and away we go." Here "high" means the numerator $f$ and "low" means the denominator $g$. The order of subtraction matters: it is $f'g - fg'$, not $fg' - f'g$. Write it out carefully on free-response problems.
Worked Example 2.10 — Product Rule Application
Differentiate $y = x^2 \sin x$.
Solution. Let $f(x) = x^2$ and $g(x) = \sin x$. Then $f'(x) = 2x$ and $g'(x) = \cos x$.
$$y' = f'g + fg' = 2x \sin x + x^2 \cos x$$
Try It 2.10 — Differentiate $y = x^3 \cos x$.
Show Hint
Let $f = x^3$, $g = \cos x$. Then $f' = 3x^2$, $g' = -\sin x$. By the product rule: $y' = 3x^2 \cos x - x^3 \sin x$.
Worked Example 2.11 — Quotient Rule Application
Differentiate $y = \dfrac{e^x}{x^2 + 1}$.
Solution. Let $f(x) = e^x$ and $g(x) = x^2 + 1$. Then $f'(x) = e^x$ and $g'(x) = 2x$.
$$y' = \frac{f'g - fg'}{g^2} = \frac{e^x(x^2+1) - e^x(2x)}{(x^2+1)^2} = \frac{e^x(x^2 - 2x + 1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2}$$
Try It 2.11 — Differentiate $y = \dfrac{\sin x}{x + 1}$.
Show Hint
Let $f = \sin x$, $g = x + 1$. Then $f' = \cos x$, $g' = 1$. By the quotient rule: $y' = \dfrac{\cos x (x+1) - \sin x}{(x+1)^2}$.
Worked Example 2.12 — Product Rule with Three Factors
Differentiate $y = x^2 e^x \cos x$.
Solution. For a product of three functions $u \cdot v \cdot w$, the product rule extends to:
$$\frac{d}{dx}[uvw] = u'vw + uv'w + uvw'$$
With $u = x^2$, $v = e^x$, $w = \cos x$; so $u' = 2x$, $v' = e^x$, $w' = -\sin x$:
$$y' = 2x \cdot e^x \cdot \cos x + x^2 \cdot e^x \cdot \cos x + x^2 \cdot e^x \cdot (-\sin x)$$
$$= xe^x(2\cos x + x\cos x - x\sin x)$$
Try It 2.12 — Differentiate $y = x \sin x \cdot e^x$.
Show Hint
Use the three-factor product rule $u'vw + uv'w + uvw'$ with $u = x$, $v = \sin x$, $w = e^x$. So $u' = 1$, $v' = \cos x$, $w' = e^x$. Result: $y' = e^x\sin x + xe^x\cos x + xe^x\sin x = e^x(\sin x + x\cos x + x\sin x)$.
2.6 Higher-Order Derivatives
Since the derivative $f'(x)$ is itself a function, we can differentiate it again. The derivative of $f'$ is called the second derivative of $f$, denoted $f''(x)$ or $\dfrac{d^2y}{dx^2}$. We can continue this process to obtain derivatives of any order.
| Order | Prime notation | Leibniz notation | Read as |
|---|---|---|---|
| First | $f'(x)$ | $\dfrac{dy}{dx}$ | "$f$ prime of $x$" |
| Second | $f''(x)$ | $\dfrac{d^2y}{dx^2}$ | "$f$ double prime of $x$" |
| Third | $f'''(x)$ | $\dfrac{d^3y}{dx^3}$ | "$f$ triple prime of $x$" |
| Fourth | $f^{(4)}(x)$ | $\dfrac{d^4y}{dx^4}$ | "$f$ to the fourth" |
| $n$-th | $f^{(n)}(x)$ | $\dfrac{d^ny}{dx^n}$ | "$f$ to the $n$-th" |
Worked Example 2.13 — All Derivatives of a Polynomial
Find all nonzero derivatives of $f(x) = 4x^3 - 2x^2 + 5x - 1$.
Solution.
$$f'(x) = 12x^2 - 4x + 5$$
$$f''(x) = 24x - 4$$
$$f'''(x) = 24$$
$$f^{(4)}(x) = 0$$
All derivatives of order 4 and higher are identically zero. This always happens for polynomials: once we differentiate enough times, the degree reaches zero and the next derivative is the zero function.
Try It 2.13 — Find all nonzero derivatives of $f(x) = x^4 - 3x^2 + 2x$.
Show Hint
$f'(x) = 4x^3 - 6x + 2$, $f''(x) = 12x^2 - 6$, $f'''(x) = 24x$, $f^{(4)}(x) = 24$, and all higher derivatives are $0$.
Physical Interpretation: Position, Velocity, and Acceleration
In mechanics, if $s(t)$ represents the position of an object at time $t$, then:
- $v(t) = s'(t)$ is the velocity (first derivative of position, signed speed).
- $a(t) = v'(t) = s''(t)$ is the acceleration (second derivative of position).
The object is at rest when $v(t) = 0$. The acceleration is positive when the velocity is increasing, negative when the velocity is decreasing.
Worked Example 2.14 — Position, Velocity, and Acceleration
A particle moves along a straight line with position $s(t) = t^3 - 6t^2 + 9t + 2$ meters, where $t$ is in seconds. Find (a) the velocity $v(t)$, (b) the acceleration $a(t)$, (c) when the particle is at rest, and (d) when the acceleration is positive.
Solution.
(a) $v(t) = s'(t) = 3t^2 - 12t + 9$
(b) $a(t) = v'(t) = 6t - 12$
(c) At rest when $v(t) = 0$: $3t^2 - 12t + 9 = 3(t-1)(t-3) = 0$, so at $t = 1$ s and $t = 3$ s.
(d) $a(t) > 0$ when $6t - 12 > 0$, i.e., $t > 2$ s.
Try It 2.14 — A particle has position $s(t) = 2t^3 - 9t^2 + 12t - 4$. Find $v(t)$ and $a(t)$, and determine when the particle is at rest.
Show Hint
$v(t) = 6t^2 - 18t + 12 = 6(t-1)(t-2)$. The particle is at rest at $t = 1$ s and $t = 2$ s. $a(t) = 12t - 18$.
AP Exam Tip: Second derivatives indicate concavity (Chapter 5). If $f''(x) > 0$, the curve is concave up (like a cup); if $f''(x) < 0$, it is concave down (like a cap). Start recognizing $f''$ as a measure of "how fast the slope is changing" — this becomes essential for curve sketching in later chapters.
Practice
Differentials
If $y = f(x)$ is differentiable, the differential $dy$ is defined by:
$$dy = f'(x)\, dx$$
Here $dx$ represents a small change in $x$, and $dy$ represents the corresponding approximate change in $y$ along the tangent line (rather than the actual change $\Delta y$ along the curve). The differential gives us a powerful tool: linear approximation.
Linear Approximation Formula
For $x$ near a point $a$ where $f$ is differentiable:
$$f(x) \approx L(x) = f(a) + f'(a)(x - a)$$
The function $L(x)$ is called the linearization of $f$ at $a$. It is simply the tangent line equation viewed as an approximation to the function near $x = a$.
Example — Approximating $\sqrt{4.1}$
Approximate $\sqrt{4.1}$ using the linearization of $f(x) = \sqrt{x}$ at $a = 4$.
Solution. We have $f(4) = 2$ and $f'(x) = \dfrac{1}{2\sqrt{x}}$, so $f'(4) = \dfrac{1}{4}$.
The linearization is $L(x) = 2 + \dfrac{1}{4}(x - 4)$. At $x = 4.1$:
$$\sqrt{4.1} \approx L(4.1) = 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025$$
The actual value is $\sqrt{4.1} \approx 2.02485...$, so the error is less than $0.00015$. The approximation is remarkably accurate for such a simple calculation.
2.7 Practice Problems
Test your understanding of the differentiation rules covered in this chapter. Each problem has a hidden solution — try to work through the problem yourself before revealing the answer.
Problem 1. Basic
Use the limit definition to find $f'(x)$ for $f(x) = 2x^2 + 3x$.
Show Solution
Compute $f(x+h) = 2(x+h)^2 + 3(x+h) = 2x^2 + 4xh + 2h^2 + 3x + 3h$.
$f(x+h) - f(x) = 4xh + 2h^2 + 3h = h(4x + 2h + 3)$.
$\dfrac{f(x+h) - f(x)}{h} = 4x + 2h + 3$.
$f'(x) = \displaystyle\lim_{h \to 0}(4x + 2h + 3) = \boxed{4x + 3}$.
Problem 2. Basic
Differentiate $y = 6x^4 - \dfrac{2}{x^3} + 5\sqrt{x} - 7$.
Show Solution
Rewrite: $y = 6x^4 - 2x^{-3} + 5x^{1/2} - 7$.
Apply the power rule term by term:
$y' = 24x^3 - 2(-3)x^{-4} + 5 \cdot \dfrac{1}{2}x^{-1/2} - 0 = \boxed{24x^3 + \dfrac{6}{x^4} + \dfrac{5}{2\sqrt{x}}}$.
Problem 3. Basic
Find the derivative of $g(t) = 3\sin t + t^2 \cos t$.
Show Solution
The first term: $\dfrac{d}{dt}[3\sin t] = 3\cos t$.
For $t^2 \cos t$, use the product rule with $u = t^2$ and $v = \cos t$:
$\dfrac{d}{dt}[t^2 \cos t] = 2t \cos t + t^2(-\sin t) = 2t\cos t - t^2 \sin t$.
Combining: $g'(t) = \boxed{3\cos t + 2t\cos t - t^2 \sin t}$.
Problem 4. Basic
Differentiate $f(x) = \dfrac{x^2 - 1}{x^2 + 1}$.
Show Solution
Apply the quotient rule with numerator $u = x^2 - 1$ and denominator $v = x^2 + 1$:
$f'(x) = \dfrac{(2x)(x^2+1) - (x^2-1)(2x)}{(x^2+1)^2} = \dfrac{2x^3 + 2x - 2x^3 + 2x}{(x^2+1)^2} = \boxed{\dfrac{4x}{(x^2+1)^2}}$.
Problem 5. Basic
Find $\dfrac{dy}{dx}$ for $y = 5e^x - 3 \cdot 4^x + \ln x$.
Show Solution
$\dfrac{dy}{dx} = 5e^x - 3 \cdot 4^x \ln 4 + \dfrac{1}{x}$.
Problem 6. Basic
Find the equation of the tangent line to $y = e^x \sin x$ at $x = 0$.
Show Solution
At $x = 0$: $y(0) = e^0 \sin 0 = 0$. The point is $(0, 0)$.
Product rule: $y' = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.
At $x = 0$: $y'(0) = e^0(\sin 0 + \cos 0) = 1(0 + 1) = 1$.
Tangent line: $y - 0 = 1(x - 0)$, so $\boxed{y = x}$.
Problem 7. Basic
Differentiate $h(x) = \dfrac{\sin x}{e^x}$.
Show Solution
Apply the quotient rule with $f = \sin x$, $g = e^x$:
$h'(x) = \dfrac{\cos x \cdot e^x - \sin x \cdot e^x}{(e^x)^2} = \dfrac{e^x(\cos x - \sin x)}{e^{2x}} = \boxed{\dfrac{\cos x - \sin x}{e^x}}$.
Problem 8. Basic
Let $f(x) = x^3 - 12x + 5$. Find all values of $x$ where the tangent line is horizontal.
Show Solution
A horizontal tangent occurs when $f'(x) = 0$.
$f'(x) = 3x^2 - 12 = 0 \implies x^2 = 4 \implies \boxed{x = \pm 2}$.
Problem 9. Basic
Differentiate $y = x \ln x - x$.
Show Solution
For $x \ln x$, use the product rule: $\dfrac{d}{dx}[x \ln x] = 1 \cdot \ln x + x \cdot \dfrac{1}{x} = \ln x + 1$.
$y' = (\ln x + 1) - 1 = \boxed{\ln x}$.
Problem 10. Basic
Find $f'(x)$ for $f(x) = \sec x \cdot \tan x$ and evaluate $f'\!\left(\dfrac{\pi}{3}\right)$.
Show Solution
Product rule with $u = \sec x$, $v = \tan x$:
$f'(x) = \sec x \tan x \cdot \tan x + \sec x \cdot \sec^2 x = \sec x \tan^2 x + \sec^3 x = \sec x(\tan^2 x + \sec^2 x)$.
At $x = \dfrac{\pi}{3}$: $\sec\dfrac{\pi}{3} = 2$, $\tan\dfrac{\pi}{3} = \sqrt{3}$, $\tan^2\dfrac{\pi}{3} = 3$, $\sec^2\dfrac{\pi}{3} = 4$.
$f'\!\left(\dfrac{\pi}{3}\right) = 2(3 + 4) = \boxed{14}$.
Problem 11. Intermediate
Find $f''(x)$ for $f(x) = 3x^5 - 2x^3 + x$.
Show Solution
$f'(x) = 15x^4 - 6x^2 + 1$.
$f''(x) = \boxed{60x^3 - 12x}$.
Problem 12. Intermediate
Find the fourth derivative $f^{(4)}(x)$ for $f(x) = x^5 - 5x^4 + 2x$.
Show Solution
$f'(x) = 5x^4 - 20x^3 + 2$.
$f''(x) = 20x^3 - 60x^2$.
$f'''(x) = 60x^2 - 120x$.
$f^{(4)}(x) = \boxed{120x - 120}$.
Problem 13. Intermediate
Consider the piecewise function $f(x) = \begin{cases} x^2 & x \leq 2 \\ 4x - 4 & x > 2 \end{cases}$. Is $f$ differentiable at $x = 2$? Justify your answer.
Show Solution
Continuity check: $\lim_{x \to 2^-} f(x) = 4$ and $\lim_{x \to 2^+} f(x) = 4(2) - 4 = 4$. Since both limits equal $f(2) = 4$, $f$ is continuous at $x = 2$.
Differentiability: Left-hand derivative: $\lim_{h \to 0^-} \dfrac{(2+h)^2 - 4}{h} = \lim_{h \to 0^-}(4 + h) = 4$.
Right-hand derivative: $\lim_{h \to 0^+} \dfrac{4(2+h) - 4 - 4}{h} = \lim_{h \to 0^+} \dfrac{4h}{h} = 4$.
Both one-sided derivatives equal 4, so $\boxed{f'(2) = 4}$ and $f$ is differentiable at $x = 2$.
Problem 14. Intermediate
Differentiate $y = x^2 \sin x \ln x$ using the three-factor product rule.
Show Solution
With $u = x^2$, $v = \sin x$, $w = \ln x$: $u' = 2x$, $v' = \cos x$, $w' = \dfrac{1}{x}$.
$y' = u'vw + uv'w + uvw'$
$= 2x \sin x \ln x + x^2 \cos x \ln x + x^2 \sin x \cdot \dfrac{1}{x}$
$= \boxed{2x \sin x \ln x + x^2 \cos x \ln x + x \sin x}$.
Problem 15. Intermediate
Differentiate $y = \dfrac{\tan x}{x^2 + 1}$.
Show Solution
Quotient rule with $f = \tan x$, $g = x^2 + 1$; $f' = \sec^2 x$, $g' = 2x$:
$y' = \dfrac{\sec^2 x \cdot (x^2+1) - \tan x \cdot 2x}{(x^2+1)^2} = \boxed{\dfrac{(x^2+1)\sec^2 x - 2x\tan x}{(x^2+1)^2}}$.
Problem 16. Intermediate
Find all $x$-values where $y = x^2 e^x$ has a horizontal tangent line.
Show Solution
Product rule: $y' = 2x e^x + x^2 e^x = xe^x(2 + x)$.
Set $y' = 0$: $xe^x(2+x) = 0$. Since $e^x > 0$ always, we need $x = 0$ or $x = -2$.
Horizontal tangent lines occur at $\boxed{x = 0}$ and $\boxed{x = -2}$.
Problem 17. Intermediate
Find the equation of the normal line (perpendicular to the tangent) to $y = x^3 - 2x$ at $x = 1$.
Show Solution
At $x = 1$: $y(1) = 1 - 2 = -1$. Point: $(1, -1)$.
$y' = 3x^2 - 2$, so $y'(1) = 1$. The tangent slope is $m_T = 1$.
The normal line is perpendicular: slope $m_N = -\dfrac{1}{m_T} = -1$.
Normal line: $y - (-1) = -1(x - 1) \Rightarrow \boxed{y = -x}$.
Problem 18. Intermediate
Show that $f(x) = |x|$ is continuous but not differentiable at $x = 0$ using limits.
Show Solution
Continuity: $\lim_{x \to 0^-}|x| = \lim_{x \to 0^-}(-x) = 0$ and $\lim_{x \to 0^+}|x| = 0$. Both one-sided limits equal $f(0) = 0$, so $f$ is continuous at $x = 0$.
Non-differentiability: Left-hand difference quotient: $\lim_{h \to 0^-}\dfrac{|0+h|-0}{h} = \lim_{h \to 0^-}\dfrac{-h}{h} = -1$.
Right-hand difference quotient: $\lim_{h \to 0^+}\dfrac{|h|}{h} = \lim_{h \to 0^+}\dfrac{h}{h} = 1$.
Since the left and right limits are $-1$ and $1$ (unequal), $f'(0)$ does not exist. $\square$
Problem 19. Intermediate
A particle moves along a straight line with position $s(t) = t^3 - 6t^2 + 9t + 2$ (meters, $t$ in seconds). Find (a) the velocity at $t = 2$, (b) the acceleration at $t = 2$, (c) when the particle is at rest, and (d) when the particle is speeding up.
Show Solution
(a) $v(t) = s'(t) = 3t^2 - 12t + 9$. At $t=2$: $v(2) = 12 - 24 + 9 = \boxed{-3}$ m/s.
(b) $a(t) = v'(t) = 6t - 12$. At $t=2$: $a(2) = 12 - 12 = \boxed{0}$ m/s$^2$.
(c) $v(t) = 0$: $3(t^2-4t+3) = 3(t-1)(t-3) = 0$. At rest at $t = 1$ s and $t = 3$ s.
(d) Speeding up when $v$ and $a$ have the same sign. Check intervals: for $t \in (0,1)$: $v > 0$, $a < 0$ (slowing); $t \in (1,2)$: $v < 0$, $a < 0$ (speeding up); $t \in (2,3)$: $v < 0$, $a > 0$ (slowing); $t > 3$: $v > 0$, $a > 0$ (speeding up). The particle speeds up on $(1, 2)$ and $(3, \infty)$.
Problem 20. Intermediate
A company's cost function is $C(x) = 0.01x^3 - 2x^2 + 100x + 500$ (dollars, $x$ = units produced). Find the marginal cost $C'(x)$ and interpret $C'(50)$.
Show Solution
$C'(x) = 0.03x^2 - 4x + 100$.
$C'(50) = 0.03(2500) - 4(50) + 100 = 75 - 200 + 100 = \boxed{-25}$.
Interpretation: When 50 units are being produced, the cost decreases by approximately $\$25$ for each additional unit produced. The negative marginal cost indicates economies of scale at this production level.
Problem 21. Challenge
Find $f''(x)$ for $f(x) = \dfrac{x^2}{x-1}$ (for $x \neq 1$).
Show Solution
First derivative (quotient rule, $u = x^2$, $v = x-1$, $u' = 2x$, $v' = 1$):
$f'(x) = \dfrac{2x(x-1) - x^2}{(x-1)^2} = \dfrac{x^2 - 2x}{(x-1)^2} = \dfrac{x(x-2)}{(x-1)^2}$.
Second derivative (quotient rule on $f'$, with $p = x^2-2x$, $q = (x-1)^2$, $p' = 2x-2$, $q' = 2(x-1)$):
$f''(x) = \dfrac{(2x-2)(x-1)^2 - (x^2-2x)\cdot 2(x-1)}{(x-1)^4}$.
Factor out $(x-1)$: $= \dfrac{(2x-2)(x-1) - 2(x^2-2x)}{(x-1)^3}$.
Expand: $(2x-2)(x-1) = 2x^2 - 4x + 2$ and $2(x^2 - 2x) = 2x^2 - 4x$.
$f''(x) = \dfrac{2x^2 - 4x + 2 - 2x^2 + 4x}{(x-1)^3} = \boxed{\dfrac{2}{(x-1)^3}}$.
Problem 22. Challenge
Prove the three-factor product rule: if $u$, $v$, $w$ are all differentiable, then $(uvw)' = u'vw + uv'w + uvw'$. Hint: Write $uvw = (uv) \cdot w$ and apply the two-factor product rule twice.
Show Solution
Let $F = uv$, so $uvw = F \cdot w$. By the two-factor product rule:
$(Fw)' = F'w + Fw'$.
Now $F' = (uv)' = u'v + uv'$ by the standard product rule.
Substituting: $(uvw)' = (u'v + uv')w + (uv)w' = u'vw + uv'w + uvw'$. $\square$
Problem 23. Challenge
Derive $\dfrac{d}{dx}[\sec x] = \sec x \tan x$ by writing $\sec x = \dfrac{1}{\cos x}$ and applying the quotient rule.
Show Solution
Write $\sec x = \dfrac{1}{\cos x}$. Using the quotient rule with $f = 1$, $g = \cos x$, $f' = 0$, $g' = -\sin x$:
$\dfrac{d}{dx}[\sec x] = \dfrac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x} = \dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x} = \sec x \tan x$. $\square$
Problem 24. Challenge
Find all $x$ where $f''(x) = 0$ for $f(x) = x^4 - 6x^2$, and verify that each is a genuine inflection point by checking the sign of $f''$ on either side.
Show Solution
$f'(x) = 4x^3 - 12x$.
$f''(x) = 12x^2 - 12 = 12(x^2 - 1)$.
Set $f''(x) = 0$: $x^2 - 1 = 0 \Rightarrow \boxed{x = \pm 1}$.
Sign analysis: For $|x| > 1$, $x^2 - 1 > 0$ so $f'' > 0$ (concave up). For $|x| < 1$, $x^2 - 1 < 0$ so $f'' < 0$ (concave down). The concavity changes at both $x = -1$ and $x = 1$, confirming they are inflection points.
Problem 25. Challenge AP-Style Multi-Part
Let $h(x) = \dfrac{x^2 e^x}{\cos x}$.
(a) Find $h'(x)$. You do not need to simplify fully.
(b) Find $h'(0)$.
(c) Write the equation of the tangent line to $h$ at $x = 0$.
(d) Find $h''(x)$ evaluated at $x = 0$.
Show Solution
(a) Write $h = \dfrac{N}{D}$ where $N = x^2 e^x$ and $D = \cos x$.
$N' = 2xe^x + x^2 e^x = xe^x(2+x)$ (product rule).
$D' = -\sin x$.
$h'(x) = \dfrac{xe^x(2+x)\cos x + x^2 e^x \sin x}{\cos^2 x} = \dfrac{xe^x[(2+x)\cos x + x\sin x]}{\cos^2 x}$.
(b) At $x = 0$: numerator $= 0 \cdot e^0[(2)\cdot 1 + 0] = 0$. Denominator $= 1$. So $h'(0) = \boxed{0}$.
(c) $h(0) = \dfrac{0 \cdot 1}{1} = 0$. The point is $(0, 0)$ and the slope is $h'(0) = 0$. Tangent line: $\boxed{y = 0}$ (the $x$-axis).
(d) To find $h''(0)$, note $h'(x) = \dfrac{g(x)}{D^2(x)}$ where $g(x) = xe^x[(2+x)\cos x + x\sin x]$. At $x=0$: $g(0) = 0$, $D^2(0) = 1$. Differentiating $g$ at $x = 0$: $g'(0) = [e^x + xe^x][(2+x)\cos x + x\sin x]|_{x=0} + xe^x[(-\cos x + (2+x)(-\sin x) + \sin x + x\cos x)]|_{x=0}$. The second term is $0$ at $x=0$. First term: $g'(0) = 1 \cdot 2 = 2$. And $(D^2)'(0) = -2\cos(0)\sin(0) = 0$. By the quotient rule: $h''(0) = \dfrac{g'(0)\cdot 1 - 0}{1^2} = \boxed{2}$.
📋 Chapter Summary
Core Definitions
$f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ — the instantaneous rate of change of $f$ at $x$.
$f$ is differentiable at $x=a$ if $f'(a)$ exists. Requires continuity, but a cusp, corner, or vertical tangent prevents differentiability.
$f''(x)$ is the derivative of $f'(x)$. In physics: position $\to$ velocity $\to$ acceleration.
$f(x) \approx f(a) + f'(a)(x-a)$ near $x=a$. Uses the tangent line to estimate nearby values.
Key Formulas
$\dfrac{d}{dx}[x^n] = nx^{n-1}$
$(fg)' = f'g + fg'$ $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$
$(\sin x)' = \cos x$ $(\cos x)' = -\sin x$ $(\tan x)' = \sec^2 x$
$(e^x)' = e^x$ $(\ln x)' = \dfrac{1}{x}$ $(a^x)' = a^x \ln a$
Differentiation Strategy
- Identify the form — product, quotient, composite, or basic function
- Apply the rule — power, product/quotient, or chain rule
- Simplify — combine like terms, factor if needed
- Check special cases — trig, $e^x$, $\ln x$ follow their own patterns