# Derivatives of cos x, sin x, e^x, ln(x)

> AP Calculus BC · Unit 2: Differentiation: Definition and Fundamental Properties
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u2-derivatives-of-cos-sin-e/

This sub-topic covers core derivative formulas for $\sin x$, $\cos x$, $e^x$, and $\ln(x)$, including their derivations, applications to tangent slopes and rate problems, and common exam traps to avoid.

**Prerequisites:** Limit definition of the derivative; Standard trigonometric limits; Properties of exponential and logarithmic functions

## Learning objectives

- Recall the derivative formulas for sin x, cos x, e^x, and ln(x)
- Derive these formulas from first principles or inverse function differentiation
- Apply the formulas to find slopes of tangents and instantaneous rates
- Avoid common sign and rule misapplication errors

## Derivatives of Sine and Cosine

The derivatives of $\sin x$ and $\cos x$ are derived directly from the limit definition of the derivative, using the two key trigonometric limits: $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$. These formulas only hold when $x$ is in radians, which is always the AP exam convention.

**Derivation:** Derive $\frac{d}{dx}[\sin x]$ from first principles

*Starting from:* Limit definition of the derivative

1. Start with the definition: $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ for $f(x) = \sin x$
2. $$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$$
3. Expand using the sine addition identity $\sin(x+h) = \sin x \cos h + \cos x \sin h$:
4. $$f'(x) = \lim_{h \to 0} \left(\sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}\right)$$
5. Apply the trigonometric limit results to evaluate the limit:
6. $$f'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$

*Conclusion:* Repeating the same process for $f(x) = \cos x$ gives $\frac{d}{dx}[\cos x] = -\sin x$.

> **Memory Cycle for Trigonometric Derivatives**
>
> Derivatives of sine and cosine cycle every four steps: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$

**Derivatives of Sine and Cosine** — For all real $x$ (in radians), $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$

*Example:* The derivative of $5\sin x$ is $5\cos x$, and the derivative of $-3\cos x$ is $3\sin x$.

**Worked example:** Find the slope of the line tangent to $f(x) = 3\sin x - 2\cos x$ at $x = \frac{\pi}{3}$. Leave your answer in exact form.

1. Apply the constant multiple and sum rules to differentiate term-by-term:
2. $$f'(x) = 3 \cdot \frac{d}{dx}[\sin x] - 2 \cdot \frac{d}{dx}[\cos x] = 3\cos x - 2(-\sin x) = 3\cos x + 2\sin x$$
3. Evaluate at $x = \frac{\pi}{3}$ using unit circle values: $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
4. $$f'\left(\frac{\pi}{3}\right) = 3\left(\frac{1}{2}\right) + 2\left(\frac{\sqrt{3}}{2}\right) = \frac{3}{2} + \sqrt{3}$$

> **Exam tip:** AP exam questions almost always require exact form for answers involving radicals and $\pi$, so never convert to a decimal unless explicitly asked.

## Derivative of the Natural Exponential Function $e^x$

The natural exponential function $f(x) = e^x$, where $e \approx 2.71828$ is Euler's number, has a unique property: it is its own derivative. This result holds for all real $x$, and only applies to the base $e$ natural exponential function.

**Derivation:** Derive $\frac{d}{dx}[e^x]$ from first principles

*Starting from:* Limit definition of the derivative

1. Start with the limit definition:
2. $$f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}$$
3. By definition, $e$ is the only positive base satisfying $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$.
4. Substitute the limit result to simplify:
5. $$f'(x) = e^x \cdot 1 = e^x$$

*Conclusion:* $\frac{d}{dx}[e^x] = e^x$ for all real $x$.

> **tip**
>
> Never misapply the power rule to $e^x$. The power rule only works when the exponent is a constant, not when the variable is in the exponent.

**Worked example:** Given $g(x) = 5e^x + 4\sin x$, find $g'(0)$, the value of the derivative at $x=0$.

1. Apply sum and constant multiple rules to differentiate term-by-term:
2. $$g'(x) = 5 \cdot \frac{d}{dx}[e^x] + 4 \cdot \frac{d}{dx}[\sin x] = 5e^x + 4\cos x$$
3. Evaluate at $x=0$, using $e^0 = 1$ and $\cos 0 = 1$:
4. $$g'(0) = 5(1) + 4(1) = 9$$

## Derivative of the Natural Logarithm $\ln(x)$

$\ln(x)$ is the inverse function of $e^x$, defined only for $x>0$. We use inverse function differentiation to derive its derivative.

**Derivation:** Derive $\frac{d}{dx}[\ln x]$ using inverse function differentiation

*Starting from:* Definition of inverse function: $y = \ln x \iff e^y = x$

1. Differentiate both sides of $e^y = x$ with respect to $x$, using the chain rule on the left:
2. $$e^y \cdot \frac{dy}{dx} = 1$$
3. Rearrange to solve for $\frac{dy}{dx}$:
4. $$\frac{dy}{dx} = \frac{1}{e^y}$$
5. Substitute back $e^y = x$ to get the result in terms of $x$:

*Conclusion:* The derivative of $\ln x$ is $\frac{1}{x}$, defined for all $x>0$. For $\ln|x|$, the derivative is also $\frac{1}{x}$ for all $x \neq 0$.

**Worked example:** Find the equation of the tangent line to $y = \ln(x) + 2e^x$ at $x=1$. Write your final answer in slope-intercept form.

1. To find a tangent line, we need the point $(1, y(1))$ and slope $y'(1)$.
2. Calculate the $y$-coordinate of the point:
3. $$y(1) = \ln(1) + 2e^1 = 0 + 2e = 2e, \text{ so the point is } (1, 2e)$$
4. Differentiate term-by-term using our core formulas:
5. $$y'(x) = \frac{1}{x} + 2e^x$$
6. Calculate the slope at $x=1$:
7. $$y'(1) = \frac{1}{1} + 2e^1 = 1 + 2e$$
8. Use point-slope form and simplify to slope-intercept:
9. $$y - 2e = (1 + 2e)(x - 1) \\ y = (1 + 2e)x - (1 + 2e) + 2e = (1 + 2e)x - 1$$

> **Exam tip:** Always check your simplification of tangent line equations: constant terms often cancel, and AP graders deduct points for incorrect final form even if the slope is correct.

## Combined AP-Style Practice

**Check your understanding**

Test your understanding with these AP-style practice problems:

1. If $f(x) = 2\cos x - 4e^x + \ln x$, which of the following is equal to $f'(x)$?

   - $-2\sin x - 4e^x + \frac{1}{x}$
   - $2\sin x - 4e^x + \ln x$
   - $-2\sin x - 4e^x + x$
   - $2\sin x + 4e^x + \frac{1}{x}$

   *Answer:* $-2\sin x - 4e^x + \frac{1}{x}$

   *Why:* Correct! You correctly applied the negative sign for cosine's derivative and the reciprocal rule for $\ln x$.

2. Let $h(x) = 3e^x - 2\sin x + \ln x$, defined for $x>0$. (a) Find $h'(x)$. (b) Find the slope of the tangent at $x = \frac{\pi}{2}$, exact form. (c) What is $h'(a)$ if $h(x)$ has a horizontal tangent at $x=a>0$?

   *Why:* This is a typical early FRQ problem that tests your recall of core derivative rules.

**Worked example:** The population of bacteria $t$ hours after the start of an experiment is modeled by $P(t) = 1000e^t + 50\sin\left(\frac{\pi t}{12}\right) + 200\ln(t+1)$ for $t \geq 0$, measured in cells. Find the instantaneous rate of change of the population at $t=0$, include units.

1. Differentiate $P(t)$ term-by-term, applying the constant multiple rule:
2. $$P'(t) = 1000e^t + 50\left(\frac{\pi}{12}\right)\cos\left(\frac{\pi t}{12}\right) + 200\left(\frac{1}{t+1}\right)$$
3. Evaluate at $t=0$, using $e^0=1$, $\cos 0=1$, $\frac{1}{0+1}=1$:
4. $$P'(0) = 1000(1) + \frac{25\pi}{6}(1) + 200(1) = 1200 + \frac{25\pi}{6} \approx 1213$$
5. The units are bacteria cells per hour. At the start of the experiment, the population is increasing at approximately 1213 cells per hour.

## Common pitfalls

- **Wrong:** Writing $\frac{d}{dx}[\sin x] = -\cos x$ or $\frac{d}{dx}[\cos x] = \sin x$, swapping derivatives and signs.
  - Why it fails: Students confuse the order of trigonometric derivatives, mixing derivative rules with antiderivative rules.
  - Correct: Write the derivative cycle $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$ on your scratch paper to confirm when needed.
- **Wrong:** Differentiating $e^x$ as $xe^{x-1}$ by incorrectly applying the power rule.
  - Why it fails: Students see an exponent and automatically use the power rule, which only works for constant exponents.
  - Correct: Check where the variable is: variable in base = power rule; variable in exponent, base $e$ = $\frac{d}{dx}[e^x] = e^x$.
- **Wrong:** Writing $\frac{d}{dx}[\ln x] = \frac{1}{x} \ln x$, adding an extra $\ln$ term.
  - Why it fails: Students mix up the derivative of $\ln x$ with other rules, or confuse it with exponential derivatives.
  - Correct: Memorize that the derivative of $\ln x$ is just the reciprocal function $\frac{1}{x}$, with no extra terms.
- **Wrong:** Differentiating $\ln(-5)$ (a constant) and getting $-\frac{1}{5}$.
  - Why it fails: Students apply the $\frac{1}{x}$ rule to any expression with $\ln$, forgetting that $\ln$ of a constant is just a constant.
  - Correct: Always check if a term is a constant before differentiating; the derivative of any constant is 0.
- **Wrong:** Writing $\frac{d}{dx}[e^x + \cos x] = e^x + \sin x$, keeping the wrong sign for cosine.
  - Why it fails: Students see the lack of a negative sign on $e^x$ or $\sin x$ and incorrectly carry the positive sign over to cosine.
  - Correct: Label the derivative of each term separately before combining to avoid sign confusion.
- **Wrong:** Writing $\frac{d}{dx}[5\ln x] = 5x$, turning the reciprocal into $x$.
  - Why it fails: Students confuse the derivative of $\ln x$ with the power rule for $x^{-1}$, or misread the problem.
  - Correct: After differentiating, double-check that $\frac{1}{x}$ does not get flipped to $x$ by mistake.

## Cheatsheet

| Function | Derivative | Notes |
| --- | --- | --- |
| $\sin x$ | $\cos x$ | All real $x$, $x$ in radians |
| $\cos x$ | $-\sin x$ | All real $x$, do not forget the negative sign |
| $e^x$ | $e^x$ | All real $x$, it is its own derivative |
| $\ln x$ | $\frac{1}{x}$ | Only defined for $x>0$, no extra terms |
| $c f(x)$ | $c f'(x)$ | Constants factor out of derivatives unchanged |
| $f(x) + g(x)$ | $f'(x) + g'(x)$ | Applies to all combinations of functions |
| Tangent at $x=a$ | $f'(a)$ | $f'(a)$ gives the slope of the tangent |

## What's next

This sub-topic gives you the four core non-polynomial derivatives that all subsequent differentiation topics build on. Correct memorization of these base formulas is critical: every more complex differentiation rule (product, quotient, chain) relies on these results, and you will reverse these rules later to find antiderivatives for integration problems. Mastery of these formulas is required for almost every question on the AP Calculus BC exam that involves differentiation.

- [The Chain Rule](https://www.owlsprep.com/study/ap-calculus-bc-u3-chain-rule/)
- [Product Rule](https://www.owlsprep.com/study/ap-calculus-bc-u2-product-rule/)
- [Quotient Rule](https://www.owlsprep.com/study/ap-calculus-bc-u2-quotient-rule/)

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