Why is the derivative of cosine negative but sine positive?

It comes from the angle-addition identities used in the limit definition. When you expand (x+h), the leading surviving term produces - x, while expanding (x+h) produces + x. Graphically, cosine is decreasing where sine is positive, which the minus sign reflects.

Do these derivative rules work if the angle is in degrees?

No. The rule d/dx[ x]= x only holds in radians, because the proof relies on _h→ 0 h/h=1, which is true only for radian measure. The AP exam always uses radians for calculus.

How is d/dx[e^x] different from d/dx[x^e]?

They use different rules. In e^x the base is constant and the exponent is the variable, so its derivative is e^x. In x^e the base is the variable and the exponent is a constant, so you apply the power rule to get ex^e-1. Read which part is the variable before choosing a rule.

Why is the domain of the derivative of ln x restricted to positive numbers?

Because x itself is only defined for x>0, its derivative 1/x is also only meaningful there. Even though 1/x is defined for negative x, that value does not represent the slope of x, which does not exist for x≤ 0.

AP-CALCBC-2.7

U2.7 Derivatives of sin, cos, eˣ, and ln x

Master the derivatives of sin x, cos x, eˣ, and ln x for AP Calculus BC — definitions, memory tricks, common errors, and worked exam problems.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U2.7 Derivatives of sin, cos, eˣ, and ln x, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U2.7 Derivatives of sin, cos, eˣ, and ln x free

What this lesson covers

Four functions show up on nearly every AP Calculus exam:

\sin x

\cos x

e^x

, and

\ln x

. Their derivatives are so common that you must recall them instantly, without deriving them each time. This lesson locks in the four rules, explains why each one makes sense, and shows how the exam disguises them inside larger problems.

You already know the power rule and linearity from earlier lessons, so here we simply add these transcendental building blocks. Once memorized, they combine with the product, quotient, and chain rules you meet next. Get these four solid now and the rest of Unit 2 becomes routine.

The Four Core Rules

Commit these to memory exactly as written:

\frac{d}{dx}[\sin x] = \cos x

\frac{d}{dx}[\cos x] = -\sin x

\frac{d}{dx}[e^x] = e^x

\frac{d}{dx}[\ln x] = \frac{1}{x}

Notice the patterns. The derivative of sine is cosine with a positive sign, but the derivative of cosine picks up a negative sign. This minus sign is the single most common source of lost points, so overlearn it.

The function

e^x

is special: it is its own derivative. No other basic function has this property, which is exactly why

e

is the natural base. Every derivative of

e^x

— first, second, hundredth — is still

e^x

.

For

\ln x

, the derivative

\frac{1}{x}

is defined only for

x > 0

, matching the domain of the natural log. The output is a rational function, not another logarithm, which surprises many students.

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\ln x$	$\frac{1}{x}$

All four assume

x

is measured in radians. Degree-based angles change the constant and are never used on the AP exam.

Why These Rules Are True

Understanding the source of each rule helps you remember it and defend it if asked.

For sine, apply the limit definition and the identity

\sin(x+h) = \sin x \cos h + \cos x \sin h

. After rearranging you rely on the two fundamental limits

\lim_{h\to 0}\frac{\sin h}{h} = 1

and

\lim_{h\to 0}\frac{\cos h - 1}{h} = 0

. These squeeze the difference quotient down to exactly

\cos x

. The same technique on

\cos x

produces

-\sin x

; the minus sign comes directly from the cosine angle-addition identity.

The rotation pattern is worth noting. Differentiating repeatedly cycles through four functions:

\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x

. Every four derivatives you return to the start. This cycle lets you find high-order derivatives quickly by taking the order modulo 4.

For

e^x

, the defining feature of the natural exponential is that its rate of change equals its current value, so the slope at every point equals the height. For

\ln x

, since

\ln x

is the inverse of

e^x

, implicit differentiation of

y = \ln x

, i.e.

e^y = x

, gives

e^y \frac{dy}{dx} = 1

, so

\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}

. You are not required to reproduce these proofs on demand, but recognizing them cements the results.

Combining With Linearity and Constants

On the exam these four rarely appear alone. Most often you differentiate a sum with constant coefficients, using the linearity rule from U2.5:

\frac{d}{dx}[a f(x) + b g(x)] = a f'(x) + b g'(x)

For example,

\frac{d}{dx}[3\sin x - 5e^x + 2\ln x] = 3\cos x - 5e^x + \frac{2}{x}

. Constants multiply straight through; each term is handled separately.

Watch the boundaries with the chain rule, which comes later. In this lesson the argument is always plain

x

. If you see

\sin(2x)

e^{3x}

, those require the chain rule and are not yet in scope — but knowing the base derivatives is the prerequisite.

A frequent trap:

\frac{d}{dx}[e^x]

e^x

, but

\frac{d}{dx}[x^e]

uses the power rule and equals

e x^{e-1}

. The exponent being a variable versus a constant completely changes the rule. Similarly,

\frac{d}{dx}[\ln x]=\frac{1}{x}

but

\frac{d}{dx}[x \ln x]

needs the product rule.

Expression	Rule to use	Derivative
$e^x$	exponential	$e^x$
$x^e$	power	$e x^{e-1}$
$\ln x$	log	$\frac{1}{x}$
$x^{-1}$	power	$-x^{-2}$

Read each term carefully and identify which rule governs it before writing anything.

How the Exam Tests These

Multiple-choice questions often bury one of these derivatives inside a larger expression, then test whether you tracked signs and coefficients. A classic distractor gives

+\sin x

instead of

-\sin x

for the derivative of cosine, or offers

\frac{1}{x}

when the correct answer required the chain rule. Always double-check the cosine sign and confirm the argument is simply

x

.

Another common item asks for a second or higher derivative. Because trig derivatives cycle, you can be asked for

f^{(50)}(x)

where

f(x)=\sin x

. Divide 50 by 4 to get remainder 2, so

f^{(50)}(x) = -\sin x

. For

e^x

, every derivative stays

e^x

, so higher-order questions there are freebies.

Free-response problems fold these into slope, tangent-line, and motion contexts. You may need

f'(a)

to build a tangent line, or evaluate a derivative at a specific

x

such as

\frac{d}{dx}[\ln x]

x=4

, giving

\frac{1}{4}

. Calculator-active sections still expect you to know the exact symbolic derivative before plugging in.

A reliable strategy: rewrite the function into clean terms, differentiate term by term, then simplify. Show each derivative explicitly, since AP readers award partial credit for a correctly stated derivative even if arithmetic slips afterward.

Key terms

Transcendental function.: A function that is not algebraic, such as $\sin x$ , $\cos x$ , $e^x$ , and $\ln x$ ; the four functions whose derivatives this lesson memorizes.
Natural exponential function.: The function $e^x$ with base $e\approx 2.718$ ; it equals its own derivative, $\frac{d}{dx}[e^x]=e^x$ .
Natural logarithm.: The inverse of $e^x$ , written $\ln x$ , defined for $x>0$ , with derivative $\frac{1}{x}$ .
Radian measure.: The angle unit required for all trig derivative rules; $\frac{d}{dx}[\sin x]=\cos x$ holds only in radians.
Linearity of the derivative.: The property that $\frac{d}{dx}[af+bg]=af'+bg'$ , letting you differentiate sums term by term with constants pulled out.
Higher-order derivative.: A derivative taken more than once; for $\sin x$ and $\cos x$ the results cycle with period 4.
Difference quotient.: The expression $\frac{f(x+h)-f(x)}{h}$ whose limit defines the derivative and proves these four rules.

Worked example

Let

f(x) = 4\sin x - 2\cos x + 3e^x - 5\ln x

. Find

f'(x)

, then evaluate

f'\left(\frac{\pi}{2}\right)

using

e^{\pi/2}\approx 4.810

Differentiate term by term using linearity. Pull each constant through and apply the core rule to each function.

The derivative of

4\sin x

4\cos x

.

The derivative of

-2\cos x

-2(-\sin x) = 2\sin x

. Be careful: the minus sign in the cosine rule combines with the coefficient's minus sign to give a positive result.

The derivative of

3e^x

3e^x

, since

e^x

is its own derivative.

The derivative of

-5\ln x

-5\cdot\frac{1}{x} = -\frac{5}{x}

.

Combining:

f'(x) = 4\cos x + 2\sin x + 3e^x - \frac{5}{x}

Now evaluate at

x=\frac{\pi}{2}

. We know

\cos\frac{\pi}{2}=0

and

\sin\frac{\pi}{2}=1

f'\left(\tfrac{\pi}{2}\right) = 4(0) + 2(1) + 3e^{\pi/2} - \frac{5}{\pi/2}

That is

0 + 2 + 3(4.810) - \frac{10}{\pi}

. Compute:

3(4.810)=14.430

and

\frac{10}{\pi}\approx 3.183

f'\left(\tfrac{\pi}{2}\right) \approx 2 + 14.430 - 3.183 = 13.247

So the slope of

f

x=\frac{\pi}{2}

is about

13.25

Practice questions

What is

\frac{d}{dx}\left[2e^x - \cos x\right]

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

Answer: $2e^x + \sin x$

The derivative of

2e^x

2e^x

because

e^x

is its own derivative. The derivative of

\cos x

-\sin x

, so the derivative of

-\cos x

-(-\sin x)=+\sin x

. Combining gives

2e^x+\sin x

. Choice with

-\sin x

drops the double-negative; the

x^{e-1}

choice wrongly uses the power rule on

e^x

f(x)=\sin x

, find

f^{(19)}(x)

, the nineteenth derivative.

Answer: $f^{(19)}(x) = -\cos x$

Derivatives of

\sin x

cycle every four steps:

\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x

. Divide 19 by 4 to get remainder 3. A remainder of 1 gives

\cos x

, 2 gives

-\sin x

, 3 gives

-\cos x

, and 0 returns to

\sin x

. Since the remainder is 3,

f^{(19)}(x)=-\cos x

Find the equation of the tangent line to

g(x)=\ln x

x=e

Answer: $y = \frac{1}{e}(x-e) + 1$ , equivalently $y = \frac{x}{e}$

First,

g(e)=\ln e = 1

, so the point is

(e,1)

. The derivative is

g'(x)=\frac{1}{x}

, so the slope at

x=e

g'(e)=\frac{1}{e}

. Point-slope form gives

y-1=\frac{1}{e}(x-e)

. Distributing,

y = \frac{x}{e} - 1 + 1 = \frac{x}{e}

, a clean simplification.

FAQ

Why is the derivative of cosine negative but sine positive?: It comes from the angle-addition identities used in the limit definition. When you expand $\cos(x+h)$ , the leading surviving term produces $-\sin x$ , while expanding $\sin(x+h)$ produces $+\cos x$ . Graphically, cosine is decreasing where sine is positive, which the minus sign reflects.
Do these derivative rules work if the angle is in degrees?: No. The rule $\frac{d}{dx}[\sin x]=\cos x$ only holds in radians, because the proof relies on $\lim_{h\to 0}\frac{\sin h}{h}=1$ , which is true only for radian measure. The AP exam always uses radians for calculus.
How is $\frac{d}{dx}[e^x]$ different from $\frac{d}{dx}[x^e]$ ?: They use different rules. In $e^x$ the base is constant and the exponent is the variable, so its derivative is $e^x$ . In $x^e$ the base is the variable and the exponent is a constant, so you apply the power rule to get $ex^{e-1}$ . Read which part is the variable before choosing a rule.
Why is the domain of the derivative of ln x restricted to positive numbers?: Because $\ln x$ itself is only defined for $x>0$ , its derivative $\frac{1}{x}$ is also only meaningful there. Even though $\frac{1}{x}$ is defined for negative $x$ , that value does not represent the slope of $\ln x$ , which does not exist for $x\le 0$ .

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