Trig, Exponential & Log Derivatives | AP Calculus 2.7

1. Four essential derivative formulas

Before using the formulas: Sine and cosine use radian inputs. The functions $\sin x$ , $\cos x$ and $e^{x}$ , and their derivatives, have all real inputs as their domains. The function $\ln x$ and its derivative here require $x > 0$ .

Core concept: Learn the derivatives of $\sin x$ , $\cos x$ , $e^{x}$ and $\ln x$ , then combine them with the power rule and linearity. Trigonometric inputs are measured in radians; the real natural logarithm requires $x > 0$ . Also recognise limits that represent derivatives of these familiar functions.

MEMORIZE THESE FOUR FORMULAS!

Function	Derivative	Memory Aid
$\sin x$	$\cos x$	Sine → Cosine (co-function)
$\cos x$	$- \sin x$	Cosine → Negative Sine
$e^{x}$	$e^{x}$	Natural exponential stays the same!
$\ln x$	$\frac{1}{x}$	Natural log → Reciprocal

These four formulas are used in many later trigonometric, exponential and logarithmic calculations.

2. Understand and justify each derivative

DERIVATIVE 1: SINE FUNCTION

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

In Words: The derivative of sine is cosine.

Why This Makes Sense:

The magnitude of the sine function's rate is greatest at $x = k π$ , where $k$ is an integer. Its value is zero there, while $\cos (k π) = (- 1)^{k}$ gives either the greatest positive rate or the greatest negative rate.
For example, the derivative at zero is $\cos 0 = 1$ , while at $π$ it is $\cos π = - 1$ . The first is increasing; the second is decreasing at that input.
At a sine maximum or minimum, $\sin x = 1$ or $\sin x = - 1$ , the derivative $\cos x = 0$ . A zero derivative describes the tangent slope at that input.

Proof Sketch (From Limit Definition):

\frac{d}{d x} [\sin x] = lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h}

Using the angle addition formula $\sin (x + h) = \sin x \cos h + \cos x \sin h$ and the special limits:

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

We get $\frac{d}{d x} [\sin x] = \cos x$

Examples:

$\frac{d}{d x} [\sin x] = \cos x$
$\frac{d}{d x} [5 \sin x] = 5 \cos x$
$\frac{d}{d x} [\sin x + x^{2}] = \cos x + 2 x$

DERIVATIVE 2: COSINE FUNCTION

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

In Words: The derivative of cosine is negative sine.

Sign check: Keep the minus sign in the derivative of cosine; losing it is a common error.

Why the Negative Sign?

At $x = 0$ , $\cos 0 = 1$ , a maximum value.
On $0 < x < π$ , cosine decreases because $- \sin x < 0$ . This is a local sign explanation, not a claim that cosine decreases everywhere.
At zero the derivative is $- \sin 0 = 0$ ; immediately to the right it is negative.
The cosine graph is decreasing immediately to the right of zero. The derivative changes sign on other intervals as sine changes sign.

Proof Sketch:

\frac{d}{d x} [\cos x] = lim_{h \to 0} \frac{\cos (x + h) - \cos x}{h}

Using $\cos (x + h) = \cos x \cos h - \sin x \sin h$ and the same special limits, we get $- \sin x$

Examples:

$\frac{d}{d x} [\cos x] = - \sin x$
$\frac{d}{d x} [3 \cos x] = - 3 \sin x$
$\frac{d}{d x} [\cos x - \sin x] = - \sin x - \cos x$
$\frac{d}{d x} [- \cos x] = - (- \sin x) = \sin x$ (double negative!)

DERIVATIVE 3: NATURAL EXPONENTIAL FUNCTION

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

In words: The derivative of the natural exponential $e^{x}$ is $e^{x}$ , so its instantaneous rate equals its current value.

Why e is Special:

$e \approx 2.71828 . . .$ is the only base where $\frac{d}{d x} [a^{x}] = a^{x}$
For a fixed real base $a > 0$ , $\frac{d}{d x} [a^{x}] = a^{x} \ln a$ . The base-one case is the constant function one, with derivative zero.
This is WHY we use e as the "natural" exponential base
The function $e^{x}$ grows at a rate proportional to its current value

Proof Sketch:

\frac{d}{d x} [e^{x}] = lim_{h \to 0} \frac{e^{x + h} - e^{x}}{h} = e^{x} lim_{h \to 0} \frac{e^{h} - 1}{h}

The fundamental exponential limit $lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$ gives the result. One way to characterise the natural base is as the positive exponential base with unit rate at zero; this limit must be established in a development of the exponential function rather than inferred from a decimal approximation.

Examples:

$\frac{d}{d x} [e^{x}] = e^{x}$
$\frac{d}{d x} [7 e^{x}] = 7 e^{x}$
$\frac{d}{d x} [e^{x} + x^{3}] = e^{x} + 3 x^{2}$
$\frac{d}{d x} [- e^{x}] = - e^{x}$

For Other Bases (General Exponential):

\frac{d}{d x} [a^{x}] = a^{x} \ln a

Example: $\frac{d}{d x} [2^{x}] = 2^{x} \ln 2$

DERIVATIVE 4: NATURAL LOGARITHM FUNCTION

\frac{d}{d x} [\ln x] = \frac{1}{x} (x > 0)

In words: The derivative of the real natural logarithm is the reciprocal of its input, on the positive-input domain.

Domain note: The real function $\ln x$ is defined only for $x > 0$ , and the derivative formula here has that same domain.

Key Properties:

As $x \to \infty$ , $\frac{1}{x} \to 0$ (derivative approaches 0—ln grows slowly)
As $x \to 0^{+}$ , $\frac{1}{x} \to \infty$ (derivative blows up—steep near origin)
The derivative is always positive for $x > 0$ (ln is always increasing)

Proof Sketch:

Inverse-function explanation: The exponential is differentiable with nonzero derivative and has inverse $\ln x$ on $x > 0$ . The inverse-function theorem ensures this inverse is differentiable. The chain-rule steps below then verify its derivative formula; the inverse and chain rules are developed in Unit 3.

e^{\ln x} \cdot \frac{d}{d x} [\ln x] = 1

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

\frac{d}{d x} [\ln x] = \frac{1}{x} ✓

Examples:

$\frac{d}{d x} [\ln x] = \frac{1}{x}$
$\frac{d}{d x} [2 \ln x] = \frac{2}{x}$
$\frac{d}{d x} [\ln x + x] = \frac{1}{x} + 1$

For Other Bases (General Logarithm):

\frac{d}{d x} [\log_{a} x] = \frac{1}{x \ln a}

Example: $\frac{d}{d x} [\log_{10} x] = \frac{1}{x \ln 10}$

Complete the trigonometric proof steps

The angle-addition identities separate each quotient into terms involving the two established special limits:

(\sin x)^{'} = \sin x lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x lim_{h \to 0} \frac{\sin h}{h} = \cos x .

(\cos x)^{'} = \cos x lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x lim_{h \to 0} \frac{\sin h}{h} = - \sin x .

Here $x$ is fixed while $h \to 0$ , so $\sin x$ and $\cos x$ are constants with respect to that limit. The limits are finite, permitting the separation. The trigonometric limit equal to one depends on radian measure.

Degrees describe a differently scaled input

If $S (t) = \sin (\frac{π t}{180})$ represents the sine of a degree-valued number $t$ , its derivative with respect to that number is

S^{'} (t) = \frac{π}{180} \cos (\frac{π t}{180}) .

The extra factor explains why degree mode cannot be substituted into the standard radian derivative formula. This scaled-input calculation previews the chain rule.

3. Compare the four derivatives

Function $f (x)$	Derivative $f^{'} (x)$	Key Point
$\sin x$	$\cos x$	Sine → Cosine
$\cos x$	$- \sin x$	Negative sign!
$e^{x}$	$e^{x}$	Derivative = itself
$\ln x$	$\frac{1}{x}$	Reciprocal of the input

Sine and cosine formulas assume radians. The real natural logarithm and its reciprocal derivative in this table apply only on $x > 0$ ; a reciprocal formula by itself does not extend the logarithm to negative inputs.

4. Six worked examples with full steps

Example 1: Basic Trig Combination

Problem: Find $f^{'} (x)$ for $f (x) = 3 \sin x - 2 \cos x$

Solution:

Apply sum/difference rule: Differentiate term-by-term
First term: $\frac{d}{d x} [3 \sin x] = 3 \cos x$
Second term: $\frac{d}{d x} [- 2 \cos x] = - 2 (- \sin x) = 2 \sin x$
Combine: $f^{'} (x) = 3 \cos x + 2 \sin x$

Answer: $f^{'} (x) = 3 \cos x + 2 \sin x$

Example 2: Exponential and Polynomial

Problem: Differentiate $g (x) = 5 e^{x} + 2 x^{3} - 7$

Solution:

Term 1: $\frac{d}{d x} [5 e^{x}] = 5 e^{x}$ (exponential stays same!)
Term 2: $\frac{d}{d x} [2 x^{3}] = 6 x^{2}$ (power rule)
Term 3: $\frac{d}{d x} [- 7] = 0$ (constant rule)
Combine: $g^{'} (x) = 5 e^{x} + 6 x^{2}$

Answer: $g^{'} (x) = 5 e^{x} + 6 x^{2}$

Example 3: Natural Logarithm with Polynomial

Problem: Find $\frac{d y}{d x}$ for $y = 4 \ln x + x^{2} - \frac{1}{x}$

Solution:

Rewrite: $y = 4 \ln x + x^{2} - x^{- 1}$
Term 1: $\frac{d}{d x} [4 \ln x] = \frac{4}{x}$
Term 2: $\frac{d}{d x} [x^{2}] = 2 x$
Term 3: $\frac{d}{d x} [- x^{- 1}] = - (- 1) x^{- 2} = x^{- 2} = \frac{1}{x^{2}}$
Combine: $\frac{d y}{d x} = \frac{4}{x} + 2 x + \frac{1}{x^{2}}$

Answer: $\frac{d y}{d x} = \frac{4}{x} + 2 x + \frac{1}{x^{2}}$

Example 4: All Four Functions Together

Problem: Find $h^{'} (x)$ for $h (x) = \sin x + \cos x + e^{x} + \ln x$

Solution:

Differentiate each term:
- $\frac{d}{d x} [\sin x] = \cos x$
- $\frac{d}{d x} [\cos x] = - \sin x$
- $\frac{d}{d x} [e^{x}] = e^{x}$
- $\frac{d}{d x} [\ln x] = \frac{1}{x}$
Combine: $h^{'} (x) = \cos x - \sin x + e^{x} + \frac{1}{x}$

Answer: $h^{'} (x) = \cos x - \sin x + e^{x} + \frac{1}{x}$

Example 5: Evaluating at a Specific Point

Problem: If $f (x) = 2 \sin x + e^{x}$ , find $f^{'} (0)$

Solution:

Find derivative: $f^{'} (x) = 2 \cos x + e^{x}$
Evaluate at $x = 0$ :
- $f^{'} (0) = 2 \cos (0) + e^{0}$
- $= 2 (1) + 1$ (remember: $\cos 0 = 1$ and $e^{0} = 1$ )
- $= 3$

Answer: $f^{'} (0) = 3$

Example 6: Mixed with Fractions

Problem: Differentiate $k (x) = \frac{\sin x}{2} + \frac{e^{x}}{3} - \frac{\ln x}{4}$

Solution:

Rewrite: $k (x) = \frac{1}{2} \sin x + \frac{1}{3} e^{x} - \frac{1}{4} \ln x$
Differentiate term-by-term:
$k^{'} (x) = \frac{1}{2} \cos x + \frac{1}{3} e^{x} - \frac{1}{4} \cdot \frac{1}{x}$
Simplify: $k^{'} (x) = \frac{\cos x}{2} + \frac{e^{x}}{3} - \frac{1}{4 x}$

Answer: $k^{'} (x) = \frac{\cos x}{2} + \frac{e^{x}}{3} - \frac{1}{4 x}$

Domain check: Examples 1, 2 and 5 allow all real inputs. Examples 3, 4 and 6 contain $\ln x$ , so both the original functions and the displayed derivatives require $x > 0$ . In Example 3, differentiating the negative reciprocal produces a positive reciprocal-square term.

5. Exponential and logarithmic functions with other bases

General exponential function: constant base

a > 0

\frac{d}{d x} [a^{x}] = a^{x} \ln a (a > 0, a \neq 1)

Examples:

$\frac{d}{d x} [2^{x}] = 2^{x} \ln 2$
$\frac{d}{d x} [10^{x}] = 10^{x} \ln 10$
$\frac{d}{d x} [3^{x}] = 3^{x} \ln 3$

Note: When $a = e$ , $\ln e = 1$ , so $\frac{d}{d x} [e^{x}] = e^{x} \cdot 1 = e^{x}$

General logarithm function: constant base

a

\frac{d}{d x} [\log_{a} x] = \frac{1}{x \ln a} (a > 0, a \neq 1, x > 0)

Examples:

$\frac{d}{d x} [\log_{10} x] = \frac{1}{x \ln 10}$
$\frac{d}{d x} [\log_{2} x] = \frac{1}{x \ln 2}$

Note: When $a = e$ , $\ln e = 1$ , so $\frac{d}{d x} [\ln x] = \frac{1}{x \cdot 1} = \frac{1}{x}$

Base conditions: The derivative formula for $a^{x}$ is valid for any fixed $a > 0$ , including $a = 1$ , which gives a constant. A real logarithm base must satisfy $a > 0$ and $a \neq 1$ , and its argument must be positive. A negative constant base does not define a real exponential function on an entire real interval in the same way.

The change-of-base identity gives $\log_{a} x = \frac{\ln x}{\ln a}$ . Because $\ln a$ is a nonzero constant, the constant multiple rule gives $\frac{1}{x \ln a}$ . If $0 < a < 1$ , then $\ln a < 0$ , so both $a^{x}$ and $\log_{a} x$ are decreasing on their real domains.

Power versus exponential: The formula $\frac{d}{d x} [x^{n}] = n x^{n - 1}$ holds for a constant exponent on its differentiability domain. The formula $\frac{d}{d x} [a^{x}] = a^{x} \ln a$ holds for a constant positive base. Identify where the variable occurs before choosing the rule.

6. Important limits behind the formulas

Essential Limits: These limits are fundamental to proving the derivative formulas:

Sine Limit:
$lim_{x \to 0} \frac{\sin x}{x} = 1$
Cosine Limit:
$lim_{x \to 0} \frac{\cos x - 1}{x} = 0$
Exponential Limit:
$lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$

The trigonometric limits use radian measure and can be established geometrically with limit arguments. The exponential limit follows from the construction of the natural exponential. Once established, these limits justify the derivative formulas and can also be recognised through the derivative definition.

7. Recognise a familiar derivative inside a limit

A second core skill in Topic 2.7 is matching a limit to the definition of a derivative. Identify the function, base input and input change before evaluating.

f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} .

The equivalent moving-input form is $f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}$ . The denominator is the change in the input, not just any expression that tends to zero.

Sine at a nonzero input

lim_{h \to 0} \frac{\sin (\frac{π}{3} + h) - \sin (\frac{π}{3})}{h} = \cos (\frac{π}{3}) = \frac{1}{2} .

This is the derivative of sine evaluated at $\frac{π}{3}$ , using radians.

Cosine at zero

lim_{x \to 0} \frac{\cos x - 1}{x} = - \sin 0 = 0 .

The numerator subtracts $\cos 0 = 1$ , so the quotient matches the derivative at zero.

The natural exponential at one

lim_{x \to 1} \frac{e^{x} - e}{x - 1} = e .

The subtracted value is $e^{1} = e$ . This is the derivative value at one, not at zero.

The natural logarithm at one

lim_{h \to 0} \frac{\ln (1 + h)}{h} = lim_{h \to 0} \frac{\ln (1 + h) - \ln 1}{h} = 1 .

The logarithm's argument is positive for all sufficiently small input changes, and $\ln 1 = 0$ , so this is $(\ln x)^{'}$ evaluated at one.

Check a scaled denominator

lim_{h \to 0} \frac{e^{2 + h} - e^{2}}{3 h} = \frac{1}{3} lim_{h \to 0} \frac{e^{2 + h} - e^{2}}{h} = \frac{e^{2}}{3} .

The input change in the numerator is $h$ ; the extra constant three must be accounted for. Direct substitution gives an indeterminate form, not a zero derivative or proof that the limit does not exist. Recognising the derivative supplies the exact value.

8. Connect derivative values with tangent lines and rates

Reuse Example 5, $f (x) = 2 \sin x + e^{x}$ . Its derivative is $f^{'} (x) = 2 \cos x + e^{x}$ . At zero, the function and derivative have different values:

f (0) = 2 \sin 0 + e^{0} = 1, f^{'} (0) = 2 \cos 0 + e^{0} = 3 .

y - f (0) = f^{'} (0) (x - 0), y = 1 + 3 x .

The curve passes through $(0, 1)$ , and its tangent has slope three. If the output measures a quantity and the input measures time, that derivative is expressed in output units per time unit. It is an instantaneous rate, distinct from an average rate over an interval.

For $g (x) = \ln x$ at one, $g (1) = 0$ and $g^{'} (1) = 1$ , so the tangent is $y = x - 1$ . At negative inputs, the real function $\ln x$ has no value or tangent even though the algebraic expression $\frac{1}{x}$ is real there.

9. Strategies, memory aids and common mistakes

Essential Tips

Memorize cold: These four derivatives must be instant recall—no hesitation!
Negative-sign alert: Among the four basic formulas on this page, cosine's derivative has an explicit minus sign: $\frac{d}{d x} [\cos x] = - \sin x$ . The numerical sign of a derivative still depends on the input.
Natural base: Among functions of the form $a^{x}$ with positive constant base, $e^{x}$ is the one whose derivative equals itself. Other functions also satisfy that equation: every constant multiple $C e^{x}$ , including the zero function, has derivative equal to itself.
Log domain: $\ln x$ only defined for $x > 0$ , so derivative too
Practice evaluating: Know special values like $\sin 0, \cos 0, e^{0}, \ln 1$
Combine with other rules: Use these WITH power rule, sum rule, etc.

Memory Devices

Mnemonic for Trig Derivatives:

"Sine goes to Cosine (co-function),
Cosine goes Negative (that's the key!)."

Pattern for Repeated Differentiation:

If you keep differentiating sine or cosine, you get a 4-step cycle:

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

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

For Exponential and Log:

"Natural exponential stays intact; the log gives the reciprocal back!"

Special Values to Memorize

Trig Values:

$\sin 0 = 0$ , $\cos 0 = 1$
$\sin (\frac{π}{2}) = 1$ , $\cos (\frac{π}{2}) = 0$
$\sin π = 0$ , $\cos π = - 1$

Exponential/Log Values:

$e^{0} = 1$
$e^{1} = e \approx 2.718$
$\ln 1 = 0$
$\ln e = 1$

Common Mistakes to Avoid

Mistake 1: Losing the minus sign in cosine's derivative. The correct identity is $\frac{d}{d x} [\cos x] = - \sin x$ .
Mistake 2: Using the power rule when the variable is in the exponent. The correct formula is $\frac{d}{d x} [e^{x}] = e^{x}$ ; the expression $x e^{x - 1}$ does not give this derivative.
Mistake 3: Taking the reciprocal of the logarithm instead of the input. The correct derivative is $\frac{d}{d x} [\ln x] = \frac{1}{x}$ for $x > 0$ , not $\frac{1}{\ln x}$ .
Mistake 4: Ignoring an inner function. For $\sin (3 x)$ , the outer sine formula alone misses the inner derivative; the answer is $3 \cos (3 x)$ . The chain rule is introduced in Unit 3, rather than Topic 2.9.
Mistake 5: Forgetting domain: $\ln x$ derivative only valid for $x > 0$
Mistake 6: Mixing up $\log_{10} x$ and $\ln x$ (different bases!)
Mistake 7: Double negative confusion: $\frac{d}{d x} [- \cos x] = - (- \sin x) = \sin x$
Mistake 8: Not using radians (trig derivatives require radian mode!)

10. Six practice problems with solutions

Differentiate the six functions or evaluate the requested derivative. Use radians for every trigonometric input and check domains.

1. Sine and cosine

$f (x) = 4 \sin x + 3 \cos x$ .

Show the trigonometric sign check

f^{'} (x) = 4 \cos x + 3 (- \sin x) = 4 \cos x - 3 \sin x .

The function and derivative are defined for all real inputs.

2. Exponential and polynomial

$g (x) = e^{x} - 2 x^{2} + 5$ .

Show the term-by-term solution

g^{'} (x) = e^{x} - 4 x + 0 = e^{x} - 4 x .

The constant contributes zero. Both domains are all real inputs.

3. Logarithm, reciprocal and root

$h (x) = \ln x + \frac{1}{x} + \sqrt{x}$ .

Show the fractions and domain check

h^{'} (x) = \frac{1}{x} - \frac{1}{x^{2}} + \frac{1}{2 \sqrt{x}}, x > 0 .

Equivalently, $h^{'} (x) = \frac{1}{x} - x^{- 2} + \frac{1}{2} x^{- \frac{1}{2}}$ . The logarithm and reciprocal restrict the original function to positive inputs, and the derivative has the same domain.

4. A negative cosine term

$k (x) = - \cos x + \sin x + e^{x}$ .

Show the double negative

k^{'} (x) = - (- \sin x) + \cos x + e^{x} .

k^{'} (x) = \sin x + \cos x + e^{x} .

The first coefficient and cosine derivative supply two negative signs. Both domains are all real inputs.

5. Base-two functions

$f (x) = 2^{x} + \log_{2} x$ .

Show the base factors

f^{'} (x) = 2^{x} \ln 2 + \frac{1}{x \ln 2}, x > 0 .

Keep the natural logarithm of the base in both formulas. The original logarithm requires positive inputs.

6. Evaluate at a special angle

For $f (x) = \sin x + \cos x$ , find $f^{'} (\frac{π}{2})$ .

Show differentiation before evaluation

f^{'} (x) = \cos x - \sin x .

f^{'} (\frac{π}{2}) = \cos (\frac{π}{2}) - \sin (\frac{π}{2}) = 0 - 1 = - 1 .

The requested value is a derivative, not the function value at that angle.

11. Write a clear AP-style response

How to write a clear AP-style solution:

Recall the basic formulas: Use them fluently in routine calculations. If the question asks for a limit definition or proof, show that reasoning instead of giving only a shortcut answer.
Correct notation: Use proper notation: $\frac{d}{d x}$ , $f^{'} (x)$ , etc.
Show the requested method: Include enough differentiation, evaluation or limit work to support your result.
Differentiate before evaluating: Find $f^{'} (x)$ , then substitute an allowed input $a$ to obtain $f^{'} (a)$ . Do not confuse the function value with the rate.
Combine with other rules: Mix these with power rule, sum rule, etc.
Know special limits: $lim_{x \to 0} \frac{\sin x}{x} = 1$ appears on exams
Use radians: All trig must be in radian mode on calculator
Domain awareness: Note when $x > 0$ for ln functions

Common FRQ Formats:

Find $f^{'} (x)$ for a specified function.
Find the tangent slope at $x = a$ by evaluating $f^{'} (a)$ .
Find the tangent equation using the point $(a, f (a))$ and slope $f^{'} (a)$ .
"Evaluate $lim_{x \to 0} \frac{\sin x}{x}$ " (know special limits)
Show a stated identity for $f^{'} (x)$ with the requested reasoning.
"Find the rate of change..." (real-world context requiring derivatives)
Solve $f^{'} (x) = 0$ on the original domain. Later applications also check inputs in that domain where a derivative fails to exist.

12. Quick reference for the four formulas

THE BIG FOUR - MEMORIZE NOW!

$\frac{d}{d x} [\sin x] = \cos x$	$\frac{d}{d x} [\cos x] = - \sin x$
$\frac{d}{d x} [e^{x}] = e^{x}$	$\frac{d}{d x} [\ln x] = \frac{1}{x}$

Bonus: $\frac{d}{d x} [a^{x}] = a^{x} \ln a$ | $\frac{d}{d x} [\log_{a} x] = \frac{1}{x \ln a}$

Special Limits: $lim_{x \to 0} \frac{\sin x}{x} = 1$ | $lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$

Trigonometric inputs use radians. Natural and general logarithms have positive real arguments; general logarithm bases satisfy $a > 0$ , $a \neq 1$ . Exponential bases are fixed and positive. Apply any composite-input formula only with its required inner derivative factor.

13. Connections and next study steps

Topic 2.7 connects to:

Topics 2.8 and 2.9: The product rule and quotient rule combine these formulas for products and quotients of differentiable functions.
Unit 3: The chain rule extends the formulas to compositions such as $\sin (3 x)$ and $e^{x^{2}}$ , with an inner derivative factor.
Unit 5: Analytical applications use these derivatives for graph behavior, critical points and optimization.
Unit 4: Motion problems with harmonic motion use trig derivatives
Unit 6: Integration connects these derivative formulas with antiderivatives and accumulation.
Unit 7: Exponential growth and decay differential equations use the proportional-rate property of exponential functions.
BC Only - Unit 10: Taylor series use repeated differentiation of these functions

Remember: The four basic identities are $(\sin x)^{'} = \cos x$ , $(\cos x)^{'} = - \sin x$ , $(e^{x})^{'} = e^{x}$ and $(\ln x)^{'} = \frac{1}{x}$ . Use radians for the trigonometric formulas and $x > 0$ for the real natural logarithm. Combine them with linearity and the power rule. Check the function's structure before using a product, quotient or chain rule, and recognise familiar derivative limits as another way to evaluate a limit.

← Topic 2.6: linearity and derivative rules Next: Topic 2.8 product rule →Review the derivative definition →All Unit 2 study notes →

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