Single-Variable-Calculus · Unit 3 · Video 1 · Interactive Practice

Specific vs. General Derivative Formulas, and the Derivative of Sine

IKey Formulas

Formula	Kind	What it does
$(u + v)' = u' + v'$	General — sum rule	Splits any sum; names no function
$(cu)' = c\,u'$	General — constant multiple	Carries a constant $c$ along untouched
$\sin(a + b) = \sin a\cos b + \cos a\sin b$	Identity	The one tool available on the sine difference quotient
$\dfrac{d}{dx}\sin x = \cos x$	Specific — new today	Rests on $\lim\limits_{\Delta x \to 0}\dfrac{\cos \Delta x - 1}{\Delta x} = 0$ and $\lim\limits_{\Delta x \to 0}\dfrac{\sin \Delta x}{\Delta x} = 1$

Key Insight: A specific formula names a function — $x^n$ , $\tfrac{1}{x}$ , $\sin x$ . A general formula names an operation and never mentions a particular function. A few of the first, worked on by the second, generate the derivative of almost everything.

IITwo Kinds of Formula, One Derivative

Two general rules take the expression apart; only the last step names a particular function.

Function Step 1

IIIKeeping a Zero Over a Zero

Every split below is algebraically the same quotient; only one leaves each piece a zero over a zero.

Split the numerator Δx 0.50

IVThe Difference Quotient of Sine

The secant slope of $\sin x$ at $a$ is heading for the height of the cosine curve at $a$ .

Base point a 1.0 Δx 2.00

💡 The two limits this result rests on, $\dfrac{\cos \Delta x - 1}{\Delta x} \to 0$ and $\dfrac{\sin \Delta x}{\Delta x} \to 1$ , are granted here; they are proved later in this lecture.

VQuiz Questions

Problem 1 · Both Kinds in One Line

Given: $f(x) = 2x^4 + 6x$ — differentiate it, then say which step needed a formula that names a particular function.

What is $f'(x)$ ?

Which step needs a specific formula?

✅ Correct! The sum rule split it, the constant-multiple rule pulled the coefficients out, and only the

x^n

formula knew anything about

x^4

itself.

❌ Not quite. The second term was left undifferentiated. Here

6x = 6 \cdot x^1

, and

\frac{d}{dx}x^1 = 1

, so it contributes

6

, not

6x

❌ Not quite.

6x

is not a constant — it is

6

times

x

, and the constant-multiple rule gives

6 \cdot 1 = 6

❌ Check the exponent step.

\frac{d}{dx}x^4 = 4x^3

, and the

2

rides along in front:

2(4x^3) = 8x^3

❌ Not quite. Splitting a sum and pulling out a constant are general rules — they work for every

u

and

v

. Only

\frac{d}{dx}x^4 = 4x^3

names a function.

Show solution

Step 1 — sum rule (general):

\frac{d}{dx}(2x^4 + 6x) = \frac{d}{dx}(2x^4) + \frac{d}{dx}(6x)

Step 2 — constant multiple (general):

= 2\,\frac{d}{dx}x^4 + 6\,\frac{d}{dx}x

Step 3 — the $x^n$ formula (specific), with $n = 4$ and $n = 1$ :

= 2(4x^3) + 6(1) = 8x^3 + 6

The first two steps never mention $x^4$ ; they would look identical for $u + v$ and $cu$ whatever $u$ and $v$ are. The third step is the only one that had to know which function it was differentiating.

Problem 2 · The Identity, the Right Way Round

Given: the difference quotient $\dfrac{\sin(x + \Delta x) - \sin x}{\Delta x}$ — expand $\sin(x + \Delta x)$ , with $x$ in the role of $a$ and $\Delta x$ in the role of $b$ .

✅ Correct! Every term is a sine times a cosine, with the two letters in opposite orders — that is the pattern to remember.

❌ That is the pairing to avoid. Like is paired with like there:

\sin\cdot\sin

and

\cos\cdot\cos

. In the true identity each term mixes one sine with one cosine.

❌ Wrong identity. That is

\cos(x + \Delta x)

. The sine version has a plus sign and mixed products.

❌ Not quite. Sine is not additive:

\sin(a + b) \ne \sin a + \sin b

. Test it with

a = b = \tfrac{\pi}{2}

, where the left side is

0

and the right side is

2

Show solution

The addition formula is

\sin(a + b) = \sin a\cos b + \cos a\sin b .

With $a = x$ and $b = \Delta x$ , keeping the substituted expression in parentheses:

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

and the whole quotient becomes

\frac{\sin x\cos \Delta x + \cos x\sin \Delta x - \sin x}{\Delta x},

with the $-\sin x$ from the definition and the denominator $\Delta x$ both carried along. Nothing has been dropped: $x$ and $\Delta x$ are simply separated now, which is the form the limit needs.

Problem 3 · Keeping a Zero Over a Zero

Given: the numerator $\sin x\cos \Delta x + \cos x\sin \Delta x - \sin x$ over $\Delta x$ — choose the split that is equal to the original quotient and leaves every piece of the form $\tfrac{0}{0}$ as $\Delta x \to 0$ .

✅ Correct! Pairing

\sin x\cos \Delta x

with

-\sin x

and factoring out

\sin x

leaves

\cos \Delta x - 1 \to 0

over

\Delta x \to 0

, and the other piece is

\sin \Delta x \to 0

over

\Delta x \to 0

❌ The first piece breaks. Its numerator

\sin x\cos \Delta x

tends to

\sin x

, not to

0

, while its denominator tends to

0

— a nonzero number over zero, which is meaningless in the limit.

❌ Both pieces break. The first numerator tends to

\sin x

and the second is

-\sin x

, each over

\Delta x \to 0

. Splitting off

\sin x/\Delta x

is exactly the move to avoid.

❌ The factors are swapped. Each piece is

\tfrac{0}{0}

, but

\sin x

is the factor that came out of

\cos \Delta x - 1

and

\cos x

is the one attached to

\sin \Delta x

. As written this would give

\sin x

, not

\cos x

Show solution

Set $\Delta x = 0$ in the original numerator: $\cos 0 = 1$ and $\sin 0 = 0$ , so it becomes $\sin x - \sin x = 0$ , over a denominator that is also $0$ . The whole quotient is $\tfrac{0}{0}$ , and every piece it is broken into must stay that way.

Only $\sin x\cos \Delta x$ fails to vanish on its own — it tends to $\sin x$ — so it must be paired with the $-\sin x$ that cancels it. Factoring $\sin x$ out of that pair:

\sin x\cos \Delta x + \cos x\sin \Delta x - \sin x = \sin x\,(\cos \Delta x - 1) + \cos x\sin \Delta x

and putting each group over its own $\Delta x$ :

\frac{\sin(x + \Delta x) - \sin x}{\Delta x} = \sin x\,\frac{\cos \Delta x - 1}{\Delta x} + \cos x\,\frac{\sin \Delta x}{\Delta x}

Both brackets are now $\tfrac{0}{0}$ , so nothing has been broken. Granting $\frac{\cos \Delta x - 1}{\Delta x} \to 0$ and $\frac{\sin \Delta x}{\Delta x} \to 1$ :

\frac{d}{dx}\sin x = \sin x \cdot 0 + \cos x \cdot 1 = \cos x

Problem 4 · The Same Method on Cosine

Given: $\cos(x + \Delta x) = \cos x\cos \Delta x - \sin x\sin \Delta x$ , so the numerator of the difference quotient of $\cos x$ is $\cos x\cos \Delta x - \sin x\sin \Delta x - \cos x$ , together with the granted limits $\dfrac{\cos \Delta x - 1}{\Delta x} \to 0$ (A) and $\dfrac{\sin \Delta x}{\Delta x} \to 1$ (B) — run the same calculation.

Which two terms must be paired?

What is $\dfrac{d}{dx}\cos x$ ?

✅ Correct! Same grouping, same two limits — and the minus sign in the cosine identity turns bracket B into

-\sin x

❌ Check which term survives. As

\Delta x \to 0

\cos x\cos \Delta x \to \cos x

while

\sin x\sin \Delta x \to 0

. Only the surviving term needs a partner, and only

-\cos x

cancels it.

❌ Not quite.

\cos x\cos \Delta x

tends to

\cos x

, not to

0

. Left on its own over

\Delta x

, it is a nonzero number over zero.

❌ Check the sign. The identity carries

-\sin x\sin \Delta x

, so bracket B enters with a minus:

\cos x \cdot 0 - \sin x \cdot 1

Show solution

Pair $\cos x\cos \Delta x$ with $-\cos x$ and factor out $\cos x$ :

\frac{\cos x\cos \Delta x - \sin x\sin \Delta x - \cos x}{\Delta x} = \cos x\,\frac{\cos \Delta x - 1}{\Delta x} - \sin x\,\frac{\sin \Delta x}{\Delta x}

Each bracket is $\tfrac{0}{0}$ , so the split is legitimate. Now apply (A) and (B):

\frac{d}{dx}\cos x = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x

The method never changed: expand with the addition formula, group so that a zero stays over a zero, then spend the two granted limits.

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