Derivatives of cos, sin, e^x, ln(x) — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Derivation and memorization of derivative rules for $\sin x$, $\cos x$, $e^x$, and natural logarithm $\ln x$, application of these rules to simple function combinations, and recognition of common exam question types for this topic.

You should already know: Limit definition of the derivative, standard trigonometric limit identities, basic properties of exponential and logarithmic functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Derivatives of cos, sin, e^x, ln(x)?

This topic establishes the core derivative rules for four of the most common non-algebraic (transcendental) functions used in AP Calculus: sine, cosine, the natural exponential function, and the natural logarithm. According to the AP Calculus AB Course and Exam Description (CED), Unit 2 (which includes this topic) makes up 10–12% of the total exam score, with this specific topic appearing in nearly every exam as both standalone problems and intermediate steps in longer multi-part questions.

This topic typically shows up in both MCQ and FRQ sections: MCQ often asks you to identify the correct derivative of a combination of these functions or calculate the slope of a tangent line at a point, while FRQ uses these derivative rules as a foundation for questions about rates of change, motion, or graph analysis. Unlike derivative rules for power functions, these rules are specific to each function type and require memorization (rooted in understanding their derivation from the limit definition of the derivative) to apply correctly on exam day. Mastery of these rules is non-negotiable, as every differentiation topic that follows builds directly on these four core formulas.

2. Derivatives of Sine and Cosine

To find the derivative of $\sin x$ from first principles, we start with the limit definition of the derivative: $$\frac{d}{dx}[\sin x]=\underset{h\to 0}{\lim }\frac{\sin (x+h)-\sin x}{h}$$ Using the sine addition identity $\sin (x+h)=\sin x\cos h+\cos x\sin h$, we split the limit into two terms: $$\underset{h\to 0}{\lim }\left(\sin x\cdot \frac{\cos h-1}{h}+\cos x\cdot \frac{\sin h}{h}\right)$$ We use the standard trigonometric limits you already know: ${\lim }_{h\to 0}\frac{\cos h-1}{h}=0$ and ${\lim }_{h\to 0}\frac{\sin h}{h}=1$. Substituting these gives $\frac{d}{dx}[\sin x]=\sin x\cdot 0+\cos x\cdot 1=\cos x$.

Repeating this process for $\cos x$ with the cosine addition identity gives the second rule. The final formulas are: $$\frac{d}{dx}[\sin x]=\cos x\frac{d}{dx}[\cos x]=-\sin x$$ Intuition for these rules: The graph of $\sin x$ has a slope of 1 at $x=0$, which matches $\cos 0=1$, and slope 0 at $x=\pi /2$, which matches $\cos (\pi /2)=0$. For $\cos x$, the slope at $x=\pi /2$ is $-1$, which matches $-\sin (\pi /2)=-1$, confirming the negative sign.

Worked Example

Find the slope of the tangent line to $f(x)=3\sin x-2\cos x$ at $x=\pi$.

1. Use the sum rule and constant multiple rule to split differentiation into individual terms.
2. Apply the derivative rules for sine and cosine: $\frac{d}{dx}[3\sin x]=3\cos x$, and $\frac{d}{dx}[-2\cos x]=-2(-\sin x)=2\sin x$.
3. Combine terms to get the derivative function: $f^{\prime }(x)=3\cos x+2\sin x$.
4. Evaluate the derivative at $x=\pi$: $f^{\prime }(\pi )=3\cos (\pi )+2\sin (\pi )=3(-1)+2(0)=-3$.
5. The slope of the tangent line at $x=\pi$ is $-3$.

Exam tip: Always double-check the sign on the derivative of cosine: the negative sign is the most commonly missed detail on trig derivative MCQ questions. If your answer is off by a negative sign, this is almost always the error.

3. Derivative of the Natural Exponential Function $e^x$

The natural exponential function $f(x)=e^x$ (where $e\approx 2.71828$ is Euler's constant) has a unique derivative property that makes it the most useful function for modeling continuous growth and decay. Again, we derive it from the limit definition: $$\frac{d}{dx}[e^x]=\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}=e^x\underset{h\to 0}{\lim }\frac{e^h-1}{h}$$ By the definition of $e$, the limit ${\lim }_{h\to 0}\frac{e^h-1}{h}=1$, so we get the simple result: $$\frac{d}{dx}[e^x]=e^x$$ This means $e^x$ is the only non-zero function that is its own derivative: the slope of the tangent line at any point equals the value of the function at that point. An important note for this unit: this rule applies only when the exponent is exactly $x$. When you learn the chain rule later, you will extend this to exponents that are functions of $x$, but for basic problems in this unit, the rule holds as written.

Worked Example

Given $f(x)=5e^x+\sin x-4$, find $f^{\prime }(0)$.

1. Apply the sum/difference rule to differentiate term by term.
2. Differentiate each term: $\frac{d}{dx}[5e^x]=5e^x$, $\frac{d}{dx}[\sin x]=\cos x$, and $\frac{d}{dx}[-4]=0$ (derivative of a constant is zero).
3. Combine terms to get $f^{\prime }(x)=5e^x+\cos x$.
4. Evaluate at $x=0$: $f^{\prime }(0)=5e^0+\cos 0=5(1)+1=6$.
5. Final result: $f^{\prime }(0)=6$.

Exam tip: Don't confuse the derivative of $e^x$ with the power rule: if you mistakenly apply the power rule to $e^x$, you'll get $x{e}^{x-1}$, which is wrong. Remember $e$ is a constant, not a variable base, so power rule does not apply here.

4. Derivative of the Natural Logarithm $\ln x$

The natural logarithm $\ln x$ is the inverse function of $e^x$, defined only for $x>0$. We can derive its derivative using implicit differentiation: let $y=\ln x$, so by definition $e^y=x$ for $x>0$. Differentiate both sides with respect to $x$: $$e^y\frac{dy}{dx}=1$$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}$, since $e^y=x$. This gives us the final rule: $$\frac{d}{dx}[\ln x]=\frac{1}{x}\text{for }x>0$$ Intuition for this rule: The graph of $\ln x$ is always increasing for $x>0$, so the derivative $\frac{1}{x}$ is always positive, which matches. As $x$ gets larger, $\ln x$ increases more slowly, so $\frac{1}{x}$ approaches 0, which matches our intuition. For $x$ close to 0, $\ln x$ increases very steeply, so $\frac{1}{x}$ becomes very large, which also matches. Always remember the domain restriction: the derivative only exists for positive $x$, just like the original function.

Worked Example

Find the equation of the tangent line to $y=\ln x+2e^x$ at the point $(1,2e)$.

1. Differentiate term by term using the rules for $\ln x$ and $e^x$: $\frac{dy}{dx}=\frac{1}{x}+2e^x$.
2. Calculate the slope at $x=1$ by evaluating the derivative: $m=\frac{1}{1}+2e^1=1+2e$.
3. Use point-slope form of a line: $y-y_1=m(x-x_1)$, where $(x_1,y_1)=(1,2e)$.
4. Substitute values and simplify to slope-intercept form: $$y-2e=(1+2e)(x-1)⟹y=(1+2e)x-1$$
5. The equation of the tangent line is $y=(1+2e)x-1$.

Exam tip: Always remember the domain restriction for $\ln x$: if a question asks for the derivative of $\ln x$ at a non-positive $x$, the derivative does not exist, because $\ln x$ itself is undefined there. This is a common trick question on MCQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Differentiating $\cos x$ to get ${\cos }^{\prime }x=\sin x$ (missing the negative sign). Why: Students mix up the order of sine and cosine derivatives, memorizing both as positive. Correct move: Always write the derivative of cosine immediately with the negative sign when you start a problem, and check that the sign matches the slope of the cosine graph at a test point like $x=\pi /2$.
• Wrong move: Applying the power rule to $e^x$ to get $(e^x{)}^{\prime }=x{e}^{x-1}$. Why: Students confuse the constant base $e$ with a variable base $x$ in power functions. Correct move: When you see $e$ as the base, immediately recall it's an exponential function, so use the rule $(e^x{)}^{\prime }=e^x$, not the power rule.
• Wrong move: Writing the derivative of $\ln x$ as $(\ln x{)}^{\prime }=x$ or $(\ln x{)}^{\prime }=\ln (1/x)$. Why: Students mix up the derivative of $\ln x$ with the derivative of $e^x$, which is itself, so they incorrectly assume $\ln x$ is its own derivative too. Correct move: Always associate the natural log derivative with reciprocal: $\ln \to 1/x$, not anything else.
• Wrong move: Attempting to evaluate the derivative of $\ln x$ at $x=0$ or $x=-2$, and getting $1/x=-1/2$ as the slope. Why: Students forget the domain of $\ln x$ is only positive $x$, so the function doesn't exist for non-positive inputs, so its derivative also doesn't exist there. Correct move: Before evaluating the derivative of $\ln x$ at a point, check that the point is in the domain of $\ln x$ (i.e., $x>0$) first.
• Wrong move: Differentiating $\ln 5$ (a constant) to get $1/5$. Why: Students see the ln and automatically apply the derivative rule for ln x, forgetting that ln 5 is a constant number, not a function of x. Correct move: Always check if the term is a constant before applying any derivative rule; the derivative of any constant is zero.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is equal to $f^{\prime }(x)$ if $f(x)=2e^x-\ln x+5\cos x$? A) $2e^x-\frac{1}{x}-5\sin x$ B) $2e^x-\frac{1}{x}+5\sin x$ C) $2e^x+\frac{1}{x}-5\sin x$ D) $2x{e}^{x-1}-\frac{1}{x}-5\sin x$

Worked Solution: We differentiate term by term using the basic rules from this chapter. First, the derivative of $2e^x$ is $2e^x$, which eliminates option D, which incorrectly uses the power rule for $e^x$. Next, the derivative of $-\ln x$ is $-1/x$, which matches all remaining options. Next, the derivative of $5\cos x$ is $5(-\sin x)=-5\sin x$. Combining all terms gives $f^{\prime }(x)=2e^x-1/x-5\sin x$, which matches option A. Correct answer: A.

Question 2 (Free Response)

Let $f(x)=\sin x+e^x-\ln x$ for $x>0$. (a) Find $f^{\prime }(x)$. (b) Find the slope of the line tangent to the graph of $f(x)$ at $x=1$. (c) Given that $f(1)=\sin (1)+e$, write the equation of the tangent line to $f(x)$ at $x=1$.

Worked Solution: (a) Differentiate term by term using the core rules: $\frac{d}{dx}[\sin x]=\cos x$, $\frac{d}{dx}[e^x]=e^x$, $\frac{d}{dx}[-\ln x]=-\frac{1}{x}$. Thus, $f^{\prime }(x)=\cos x+e^x-\frac{1}{x}$ for $x>0$.

(b) The slope at $x=1$ is $f^{\prime }(1)$, so substitute $x=1$ into the derivative: $f^{\prime }(1)=\cos (1)+e^1-\frac{1}{1}=e+\cos (1)-1$. The slope of the tangent line at $x=1$ is $e+\cos (1)-1$.

(c) Use point-slope form $y-f(1)=f^{\prime }(1)(x-1)$, substitute the given and calculated values: $y-(\sin 1+e)=(e+\cos 1-1)(x-1)$. Simplifying to slope-intercept form gives: $y=(e+\cos 1-1)x+(1-\cos 1+\sin 1)$. Either form is acceptable on the AP exam.

Question 3 (Application / Real-World Style)

The number of bacteria in a petri dish $t$ hours after the start of an experiment is modeled by $N(t)=1000e^t+200\sin (\pi t)$, where $N(t)$ is measured in individual bacteria cells. Find the instantaneous rate of change of the bacteria population at $t=2$ hours. Explain what your result means in the context of the experiment.

Worked Solution: The instantaneous rate of change at $t=2$ is given by $N^{\prime }(2)$. First, differentiate $N(t)$ term by term: $N^{\prime }(t)=1000\frac{d}{dt}[e^t]+200\frac{d}{dt}[\sin (\pi t)]=1000e^t+200\pi \cos (\pi t)$ Evaluate at $t=2$: $N^{\prime }(2)=1000e^2+200\pi \cos (2\pi )=1000e^2+200\pi (1)\approx 8017$ Interpretation: At 2 hours after the start of the experiment, the bacteria population is increasing at a rate of approximately 8017 bacteria per hour.

7. Quick Reference Cheatsheet

8. What's Next

This topic gives you the core derivative rules for the four most common transcendental functions, which you will use in every upcoming differentiation topic in AP Calculus. Immediately next, you will learn the product rule and quotient rule for differentiating products and quotients of these basic functions, followed by the chain rule for differentiating composite functions like $e^{2x}$ or $\sin (x^2)$. Without memorizing these four basic derivative rules correctly, you will not be able to correctly apply product, quotient, or chain rule, leading to unnecessary errors on every subsequent differentiation problem. In the bigger picture, these rules are foundational for all later topics including graph analysis, related rates, optimization, and integration. Follow-on topics that build directly on this chapter are: Product and Quotient Rules Chain Rule Derivatives of General Exponential and Logarithmic Functions Implicit Differentiation