The Central Idea
From the previous lesson, the difference quotient is
\[ \frac{f(x+h)-f(x)}{h},\qquad h\ne0. \]It gives the average rate of change from the input \(x\) to the nearby input \(x+h\).
Differential analysis asks what happens as the input change \(h\) becomes arbitrarily small:
\[ \boxed{f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}}, \]provided that this limit exists as a finite real number.
This limit is the derivative of \(f\) at \(x\).
From Average Change to Instantaneous Change
For a nonzero interval, a rate of change compares two distinct inputs. For example,
\[ \frac{f(a+h)-f(a)}{h} \]compares the output at \(a\) with the output at \(a+h\).
As \(h\to0\), the second input approaches the first. The interval does not become zero inside the quotient; instead, we study the values of the quotient for smaller and smaller nonzero values of \(h\).
If those average rates approach one number, that number is interpreted as the instantaneous rate of change at the starting input.
The Derivative at a Specific Point
At a fixed input \(x=a\), the derivative is defined by
\[ \boxed{f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}}. \]This form is especially useful when the question asks for the derivative at one particular input.
The output \(f'(a)\), when it exists, is a number. It tells us the local rate of change of \(f\) at \(a\).
The Derivative as a Function
If we leave the starting input variable, we can define a new function:
\[ \boxed{f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}}. \]The function \(f'\) assigns to each differentiable input \(x\) the local rate of change of \(f\) there.
The original function tells us the value of a quantity. The derivative function tells us how that value is changing.
The Geometric View: Secant Lines Become Tangent Lines
The difference quotient
\[ \frac{f(a+h)-f(a)}{h} \]is the slope of the secant line through the points
\[ (a,f(a)) \]and
\[ (a+h,f(a+h)). \]As \(h\to0\), the second point moves toward the first. When the secant slopes approach a finite limit, that limiting slope is the slope of the tangent line at \(x=a\):
\[ \boxed{m_{\text{tangent}}=f'(a)}. \]A Tangent Line Is a Local Linear Approximation
If \(f'(a)\) exists, the tangent line at \(x=a\) has slope \(f'(a)\) and passes through \((a,f(a))\).
Its equation is
\[ \boxed{y-f(a)=f'(a)(x-a)}. \]Near \(x=a\), this line often approximates the graph of \(f\). Later, this observation will become the basis of linearization and differentials.
Why We Do Not Substitute \(h=0\) First
The defining quotient is
\[ \frac{f(a+h)-f(a)}{h}. \]Direct substitution of \(h=0\) produces
\[ \frac{f(a)-f(a)}{0}=\frac{0}{0}, \]which is undefined.
The derivative therefore depends on the limit of the quotient for nearby nonzero \(h\), not on the value of the original quotient at \(h=0\).
The Form \(0/0\) Is a Signal to Simplify
In many derivative limits, direct substitution produces the indeterminate form
\[ \frac{0}{0}. \]This does not mean the limit is zero, and it does not mean the limit fails to exist. It means the unsimplified quotient does not yet reveal the limiting behavior.
The usual first-principles strategy is:
- form the difference quotient;
- simplify it for \(h\ne0\);
- cancel any factor causing the removable zero denominator;
- then evaluate the limit as \(h\to0\).
Example: Derivative of \(f(x)=x^2\) at \(x=3\)
Begin with the pointwise definition:
\[ f'(3)=\lim_{h\to0}\frac{f(3+h)-f(3)}{h}. \]Since \(f(x)=x^2\),
\[ f(3+h)=(3+h)^2 \]and
\[ f(3)=9. \]Substitute:
\[ f'(3)=\lim_{h\to0}\frac{(3+h)^2-9}{h}. \]Expand:
\[ (3+h)^2=9+6h+h^2. \]Therefore,
\[ f'(3)=\lim_{h\to0}\frac{6h+h^2}{h}. \]For \(h\ne0\), factor and cancel \(h\):
\[ \frac{h(6+h)}{h}=6+h. \]Now take the limit:
\[ \boxed{f'(3)=\lim_{h\to0}(6+h)=6}. \]What the Answer \(f'(3)=6\) Means
The result
\[ f'(3)=6 \]has two immediate interpretations.
- The graph of \(y=x^2\) has tangent slope \(6\) at \(x=3\).
- The function is changing locally at about \(6\) output units per input unit near \(x=3\).
Because \(f(3)=9\), the tangent line is
\[ y-9=6(x-3). \]Example: Derivative of \(f(x)=x^2\) for General \(x\)
Now keep the starting input variable:
\[ f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}. \]Expand the square:
\[ (x+h)^2=x^2+2xh+h^2. \]Then
\[ f'(x)=\lim_{h\to0}\frac{2xh+h^2}{h}. \]Factor and cancel \(h\):
\[ f'(x)=\lim_{h\to0}(2x+h). \]Therefore,
\[ \boxed{f'(x)=2x}. \]We have produced an entire derivative function from the limit definition.
The Derivative Function Encodes a Field of Slopes
For
\[ f(x)=x^2, \]we found
\[ f'(x)=2x. \]So the tangent slope depends on location:
\[ f'(-2)=-4, \] \[ f'(0)=0, \] \[ f'(3)=6. \]The derivative function records how the original curve is changing at every differentiable input.
Example: A Linear Function
Let
\[ f(x)=mx+b. \]Then
\[ f(x+h)=m(x+h)+b=mx+mh+b. \]Form the difference quotient:
\[ \frac{f(x+h)-f(x)}{h} =\frac{mx+mh+b-(mx+b)}{h}. \]Simplify:
\[ \frac{mh}{h}=m. \]Therefore,
\[ \boxed{f'(x)=m}. \]A line has the same slope everywhere, so its derivative is constant.
Example: A Constant Function
Let
\[ f(x)=c. \]Then
\[ f(x+h)-f(x)=c-c=0. \]Hence
\[ f'(x)=\lim_{h\to0}\frac{0}{h}=0. \]Therefore,
\[ \boxed{f'(x)=0}. \]A constant function does not change, and its graph is horizontal.
Example: Derivative of \(f(x)=x^3\) from the Definition
Start with
\[ f'(x)=\lim_{h\to0}\frac{(x+h)^3-x^3}{h}. \]Expand:
\[ (x+h)^3=x^3+3x^2h+3xh^2+h^3. \]Subtract \(x^3\):
\[ 3x^2h+3xh^2+h^3. \]Divide by \(h\):
\[ 3x^2+3xh+h^2. \]Now let \(h\to0\):
\[ \boxed{f'(x)=3x^2}. \]Example: A Reciprocal Function
Let
\[ f(x)=\frac{1}{x},\qquad x\ne0. \]The difference quotient is
\[ \frac{\frac{1}{x+h}-\frac{1}{x}}{h}. \]Combine the numerator over a common denominator:
\[ \frac{1}{x+h}-\frac{1}{x} =\frac{x-(x+h)}{x(x+h)} =\frac{-h}{x(x+h)}. \]Therefore,
\[ \frac{\frac{-h}{x(x+h)}}{h} =-\frac{1}{x(x+h)}. \]Taking the limit gives
\[ \boxed{f'(x)=-\frac{1}{x^2}},\qquad x\ne0. \]Example: A Square-Root Function
Let
\[ f(x)=\sqrt{x},\qquad x>0. \]Begin with
\[ f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}. \]Multiply by the conjugate:
\[ \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}. \]The numerator becomes
\[ (x+h)-x=h. \]So, for \(h\ne0\),
\[ \frac{1}{\sqrt{x+h}+\sqrt{x}}. \]Taking the limit yields
\[ \boxed{f'(x)=\frac{1}{2\sqrt{x}}},\qquad x>0. \]An Equivalent Limit Definition
There is another common way to define the derivative at \(x=a\):
\[ \boxed{f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}. \]This is equivalent to the \(h\)-form because if
\[ h=x-a, \]then \(x=a+h\), and \(x\to a\) corresponds to \(h\to0\).
Choosing Between the Two Definitions
The two standard forms are
\[ f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} \]and
\[ f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}. \]The \(h\)-form emphasizes a shrinking input increment. The \(x\to a\) form emphasizes the slope through a fixed point and a moving point. They express the same derivative when the limit exists.
Derivative Notation
Several notations are commonly used for derivatives. If \(y=f(x)\), then the first derivative may be written as
\[ f'(x), \] \[ y', \] \[ \frac{dy}{dx}, \]or, when the function name must be explicit,
\[ \frac{d}{dx}f(x). \]These notations emphasize different viewpoints, but at this stage they all refer to the same first-derivative concept.
What \(\frac{dy}{dx}\) Means Here
The notation
\[ \frac{dy}{dx} \]resembles a fraction and is extremely useful later. In the present definition, however, it represents the derivative produced by a limit:
\[ \frac{dy}{dx} =\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}, \]when that limit exists.
The finite change ratio \(\Delta y/\Delta x\) and the derivative \(dy/dx\) are related but not identical objects.
Derivative Values Have Units
If \(f(x)\) has output units and \(x\) has input units, then
\[ \frac{f(x+h)-f(x)}{h} \]has units
\[ \frac{\text{output units}}{\text{input units}}. \]The derivative has the same rate units. For example, if position is measured in meters and time in seconds, then its derivative has units of meters per second.
Application: Instantaneous Velocity
Suppose \(s(t)\) gives position at time \(t\). Average velocity over the interval from \(t\) to \(t+h\) is
\[ \frac{s(t+h)-s(t)}{h}. \]Instantaneous velocity is the limiting average velocity:
\[ \boxed{v(t)=s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}}. \]The derivative therefore converts a position function into a velocity function.
Example: Velocity from a Position Function
Let
\[ s(t)=5t^2. \]Then
\[ s(t+h)=5(t+h)^2=5t^2+10th+5h^2. \]The difference quotient is
\[ \frac{10th+5h^2}{h}=10t+5h. \]Taking \(h\to0\),
\[ \boxed{v(t)=s'(t)=10t}. \]At \(t=3\), the instantaneous velocity is
\[ \boxed{v(3)=30}. \]Application: Marginal Change
Suppose \(C(q)\) is a cost function, where \(q\) is quantity produced. The derivative
\[ C'(q) \]measures the local rate at which cost changes with output.
In applications this is called marginal cost. It arises from the same limit structure:
\[ C'(q)=\lim_{h\to0}\frac{C(q+h)-C(q)}{h}. \]What Does It Mean for a Function to Be Differentiable?
A function is differentiable at \(x=a\) if the derivative
\[ f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} \]exists as a finite real number.
A function is differentiable on an interval if it is differentiable at the relevant points of that interval, with one-sided interpretation at endpoints when appropriate.
One-Sided Derivatives
Just as limits can be taken from one side, derivative limits can be taken from one side.
The right-hand derivative at \(a\) is
\[ f'_+(a)=\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}, \]and the left-hand derivative is
\[ f'_-(a)=\lim_{h\to0^-}\frac{f(a+h)-f(a)}{h}. \]At an interior point, the ordinary derivative exists only when both one-sided derivatives exist and are equal.
Example: A Corner at \(x=0\)
Consider
\[ f(x)=|x|. \]Since \(f(0)=0\), the difference quotient at \(0\) is
\[ \frac{|h|}{h}. \]If \(h>0\), then \(|h|=h\), so
\[ \frac{|h|}{h}=1. \]If \(h<0\), then \(|h|=-h\), so
\[ \frac{|h|}{h}=-1. \]Therefore,
\[ f'_+(0)=1, \qquad f'_-(0)=-1. \]The one-sided derivatives disagree, so
\[ \boxed{f'(0)\text{ does not exist}}. \]Why Corners Break Differentiability
At a smooth point, secant slopes from the left and right approach the same tangent slope.
At a corner, the graph approaches the point with different limiting slopes from opposite sides. There is no single finite slope that represents the local direction of the curve.
Cusps Can Also Break Differentiability
A cusp is a sharp point where slopes may become unbounded with incompatible behavior from opposite sides.
The key question is still the defining limit: does
\[ \frac{f(a+h)-f(a)}{h} \]approach one finite real number as \(h\to0\)? If not, the ordinary derivative does not exist there.
Vertical Tangents and Infinite Slope Behavior
Sometimes the difference quotient grows without bound as \(h\to0\). A graph may then have a vertical tangent in an extended geometric sense.
Under the ordinary real-valued derivative definition used here, an infinite limit is not a finite derivative value. We therefore say the ordinary derivative does not exist as a finite real number at that point.
Discontinuities Break Differentiability
A function cannot be differentiable at an interior point where it is discontinuous.
Intuitively, a derivative requires nearby function values to fit together closely enough to produce one limiting local slope. A jump, hole not filled by the correct value, or other discontinuity prevents that smooth local behavior.
Differentiability Implies Continuity
If \(f'(a)\) exists, then \(f\) must be continuous at \(a\).
One way to see the connection is to write
\[ f(a+h)-f(a) =h\left(\frac{f(a+h)-f(a)}{h}\right). \]If the quotient approaches the finite number \(f'(a)\) while \(h\to0\), then the product approaches \(0\). Thus
\[ f(a+h)\to f(a). \]That is continuity at \(a\).
Continuity Does Not Guarantee Differentiability
The converse is false. A function may be continuous at a point but still fail to have a derivative there.
The function
\[ f(x)=|x| \]is continuous at \(0\), but the left and right derivative limits are different. Continuity keeps the graph connected locally; differentiability demands the stronger condition of one finite local slope.
A Practical Differentiability Checklist
When examining a point \(x=a\), ask:
- Is \(f(a)\) defined?
- Is the function continuous at \(a\)?
- Do the nearby secant slopes approach finite values?
- Do the left-hand and right-hand derivative limits agree?
If the answer to all relevant questions is yes, the function is differentiable there.
Numerical Evidence for a Derivative
A derivative can often be estimated numerically by evaluating the difference quotient for small positive and negative values of \(h\).
For \(f(x)=x^2\) at \(x=3\), the quotient simplifies to
\[ 6+h. \]Then
\[ h=0.1\quad\Rightarrow\quad6.1, \] \[ h=0.01\quad\Rightarrow\quad6.01, \] \[ h=-0.01\quad\Rightarrow\quad5.99, \] \[ h=-0.001\quad\Rightarrow\quad5.999. \]These values support the exact result \(f'(3)=6\).
Numerical Evidence Is Not the Same as an Exact Limit
A table of shrinking \(h\)-values can strongly suggest a derivative, but a finite list of numerical values does not by itself prove the exact limit.
When algebraic simplification is available, the limit calculation gives an exact result. Numerical estimates are useful for intuition, checking, and cases where symbolic work is difficult.
The Derivative Is Local Information
The value \(f'(a)\) depends on how \(f\) behaves arbitrarily close to \(a\). It does not require knowing the behavior of the function far away.
This local character is one reason derivatives are so useful: they convert a complicated global curve into a collection of local rates and local linear approximations.
The Sign of the Derivative
The derivative carries directional information.
- If \(f'(a)>0\), the graph has positive instantaneous slope at \(a\).
- If \(f'(a)<0\), the graph has negative instantaneous slope at \(a\).
- If \(f'(a)=0\), the tangent is horizontal.
To conclude that a function is increasing or decreasing across an interval, we examine the sign of the derivative throughout that interval rather than at only one point. A zero derivative also does not automatically mean a maximum or minimum. Those questions are developed later in Differential Analysis.
Common Error: Treating the Difference Quotient as the Derivative
The expression
\[ \frac{f(x+h)-f(x)}{h} \]with nonzero \(h\) is an average rate of change. It becomes a derivative only after taking the limit as \(h\to0\):
\[ \boxed{f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}}. \]Common Error: Substituting \(h=0\) Before Simplifying
For \(f(x)=x^2\), direct substitution into
\[ \frac{(x+h)^2-x^2}{h} \]gives \(0/0\).
The correct approach is to simplify for \(h\ne0\):
\[ \frac{2xh+h^2}{h}=2x+h, \]and then take the limit.
Common Error: Canceling a Term Instead of a Factor
From
\[ \frac{2xh+h^2}{h}, \]we may factor the numerator:
\[ \frac{h(2x+h)}{h}. \]Now the common factor \(h\) can be canceled for \(h\ne0\).
Cancellation works with multiplicative factors, not across addition.
Common Error: Confusing \(f'(a)\) with \(f(a)\)
The value \(f(a)\) is the output of the original function. The value \(f'(a)\) is the local rate of change of the function at that input.
For \(f(x)=x^2\) at \(x=3\),
\[ f(3)=9, \]while
\[ f'(3)=6. \]These numbers answer different questions.
Common Error: Assuming Every Continuous Point Is Differentiable
Continuity is necessary for differentiability at an interior point, but it is not sufficient.
The graph of \(y=|x|\) is continuous at \(0\), yet its left and right slopes disagree. A corner can therefore be continuous without being differentiable.
Common Error: Ignoring One Side of the Limit
At an interior point, checking only \(h>0\) or only \(h<0\) is not enough. The derivative requires a two-sided limit.
If
\[ f'_-(a)\ne f'_+(a), \]then the ordinary derivative at \(a\) does not exist.
Common Error: Forgetting the Original Domain
Algebraic simplification can produce a cleaner expression, but it does not erase the original domain restrictions.
For
\[ f(x)=\frac{1}{x}, \]the derivative formula
\[ -\frac{1}{x^2} \]is valid only where the original function is defined, namely \(x\ne0\).
A Reliable Procedure: Derivative at a Point
- Identify the function \(f\) and the input \(a\).
- Write \(\displaystyle f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\).
- Compute \(f(a+h)\) carefully.
- Subtract \(f(a)\) with parentheses when needed.
- Simplify the numerator.
- Factor, combine fractions, or rationalize as appropriate.
- Cancel only valid common factors for \(h\ne0\).
- Evaluate the remaining limit.
- Interpret the result as a local rate or tangent slope.
A Reliable Procedure: Derivative Function from First Principles
- Write \(\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\).
- Substitute \(x+h\) everywhere \(x\) appears in the function.
- Simplify the difference \(f(x+h)-f(x)\).
- Expose a factor of \(h\) when possible.
- Cancel the factor \(h\) for nonzero \(h\).
- Take the limit as \(h\to0\).
- State any domain restrictions on the resulting derivative.
Example: Full Pointwise Workflow
Let
\[ f(x)=x^2+4x \]and find \(f'(2)\) from the limit definition.
First,
\[ f(2)=4+8=12. \]Next,
\[ f(2+h)=(2+h)^2+4(2+h). \]Expand:
\[ f(2+h)=4+4h+h^2+8+4h=12+8h+h^2. \]Form the quotient:
\[ \frac{f(2+h)-f(2)}{h} =\frac{8h+h^2}{h} =8+h. \]Therefore,
\[ \boxed{f'(2)=\lim_{h\to0}(8+h)=8}. \]Example: Full Derivative-Function Workflow
Let
\[ f(x)=2x^2-3x+5. \]Then
\[ f(x+h)=2(x+h)^2-3(x+h)+5. \]Expand:
\[ f(x+h)=2x^2+4xh+2h^2-3x-3h+5. \]Subtract \(f(x)\):
\[ f(x+h)-f(x)=4xh+2h^2-3h. \]Divide by \(h\):
\[ 4x+2h-3. \]Take the limit:
\[ \boxed{f'(x)=4x-3}. \]Practice 1: Derivative at a Point
Use the limit definition to find \(f'(2)\) for
\[ f(x)=x^2. \]Practice 2: Derivative of a Linear Function
Use the limit definition to find \(f'(x)\) for
\[ f(x)=7x-4. \]Practice 3: Derivative of a Quadratic
Use first principles to find \(f'(x)\) for
\[ f(x)=3x^2+2x-1. \]Practice 4: Constant Function
Use the limit definition to find the derivative of
\[ f(x)=12. \]Practice 5: Cubic at a Point
Given
\[ f(x)=x^3, \]use the limit definition to find \(f'(2)\).
Practice 6: Tangent Line
For
\[ f(x)=x^2, \]use the derivative from limits to find the tangent line at \(x=1\).
Practice 7: One-Sided Derivatives
For
\[ f(x)=|x|, \]state \(f'_-(0)\), \(f'_+(0)\), and whether \(f'(0)\) exists.
Practice 8: Interpret the Units
A function \(T(t)\) measures temperature in degrees Celsius, where \(t\) is measured in hours. What units does \(T'(t)\) have?
Practice 9: Instantaneous Velocity
A particle has position
\[ s(t)=3t^2+2t. \]Use the limit definition to find the velocity function.
Practice 10: Identify the Error
A student wants the derivative of \(f(x)=x^2\) and writes
\[ f'(x)=\frac{(x+0)^2-x^2}{0}=\frac{0}{0}. \]The student concludes that the derivative does not exist. Explain the error.
Practice Answers
Practice 1
\[ f'(2)=\lim_{h\to0}\frac{(2+h)^2-4}{h} =\lim_{h\to0}(4+h) =\boxed{4}. \]Practice 2
\[ \frac{7(x+h)-4-(7x-4)}{h} =\frac{7h}{h}=7. \] \[ \boxed{f'(x)=7}. \]Practice 3
\[ f(x+h)=3(x+h)^2+2(x+h)-1. \]After expansion and subtraction,
\[ f(x+h)-f(x)=6xh+3h^2+2h. \]Therefore,
\[ \frac{f(x+h)-f(x)}{h}=6x+3h+2, \]so
\[ \boxed{f'(x)=6x+2}. \]Practice 4
\[ \boxed{f'(x)=0}. \]Practice 5
Since the first-principles calculation gives
\[ f'(x)=3x^2, \]we have
\[ \boxed{f'(2)=12}. \]Practice 6
For \(f(x)=x^2\),
\[ f'(x)=2x. \]At \(x=1\), the point is \((1,1)\) and the slope is \(2\). Therefore,
\[ y-1=2(x-1), \]or
\[ \boxed{y=2x-1}. \]Practice 7
\[ \boxed{f'_-(0)=-1}, \] \[ \boxed{f'_+(0)=1}. \]Because the one-sided derivatives are unequal,
\[ \boxed{f'(0)\text{ does not exist}}. \]Practice 8
The derivative measures temperature change per unit time, so its units are
\[ \boxed{\text{degrees Celsius per hour}}. \]Practice 9
\[ s(t+h)=3(t+h)^2+2(t+h). \]Subtracting \(s(t)\) gives
\[ 6th+3h^2+2h. \]Divide by \(h\):
\[ 6t+3h+2. \]Therefore,
\[ \boxed{v(t)=s'(t)=6t+2}. \]Practice 10
The student substituted \(h=0\) before taking the limit. The derivative is defined by the behavior of the quotient as \(h\) approaches zero through nonzero values. The quotient must first be simplified:
\[ \frac{(x+h)^2-x^2}{h} =\frac{2xh+h^2}{h} =2x+h. \]Then
\[ \boxed{f'(x)=\lim_{h\to0}(2x+h)=2x}. \]What to Remember
The derivative at a point is defined by
\[ \boxed{f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}}, \]and the derivative function is
\[ \boxed{f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}}, \]wherever the limit exists as a finite real number.
The derivative represents:
- instantaneous rate of change;
- slope of the tangent line;
- the limit of nearby secant slopes;
- local sensitivity of output to input.
The central operational habit is equally important: do not set the input increment equal to zero inside the original difference quotient. Simplify for nonzero increments, then take the limit.
Where We Go Next
The limit definition explains what a derivative is. Repeating a first-principles limit calculation every time, however, would be inefficient for complicated expressions.
The next lesson, Derivative Rules, develops reusable rules for constants, sums, scalar multiples, powers, and other common forms. Those rules are consequences of the derivative definition and limit laws; they do not replace the underlying meaning established here.