The Central Idea
From the limit definition,
\[ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}. \]This definition is fundamental, but using it directly for every function would be slow. Once recurring patterns have been established from the limit definition, we can use them as derivative rules.
The core idea is:
\[ \boxed{\text{understand the derivative through limits, then compute it efficiently with rules.}} \]What Differentiation Does
Differentiation is the process of finding a derivative. If
\[ y=f(x), \]differentiation produces a new function
\[ y'=f'(x). \]The original function reports output values. The derivative reports how rapidly those outputs are changing with respect to the input.
Derivative Notation
Several notations may describe the same derivative. If
\[ y=f(x), \]common forms include
\[ f'(x),\qquad y',\qquad \frac{dy}{dx},\qquad \frac{d}{dx}[f(x)]. \]The operator notation
\[ \frac{d}{dx}[\,\cdot\,] \]is especially useful when applying rules because it explicitly says: differentiate the expression inside with respect to \(x\).
The Constant Rule
A constant does not change as \(x\) changes. Therefore, its rate of change is zero.
If \(c\) is a constant, then
\[ \boxed{\frac{d}{dx}[c]=0}. \]For example,
\[ \frac{d}{dx}[7]=0, \]and
\[ \frac{d}{dx}[-\pi]=0. \]Why the Constant Rule Works
Let
\[ f(x)=c. \]Then
\[ f(x+h)=c. \]The difference quotient becomes
\[ \frac{f(x+h)-f(x)}{h} =\frac{c-c}{h} =0. \]Therefore,
\[ \boxed{f'(x)=0}. \]The rule is not arbitrary. It is a compact result obtained from the limit definition.
The Identity Function
The simplest changing function is
\[ f(x)=x. \]Its graph is a line with slope \(1\), so its derivative is
\[ \boxed{\frac{d}{dx}[x]=1}. \]This result is also the \(n=1\) case of the power rule introduced next.
The Power Rule
For a power of \(x\), differentiation follows one of the most useful patterns in elementary calculus.
For appropriate real exponents \(n\) on intervals where the power function is defined and differentiable,
\[ \boxed{\frac{d}{dx}[x^n]=nx^{n-1}}. \]In words:
- bring the exponent down as a coefficient;
- subtract \(1\) from the exponent.
Power Rule Example: A Positive Integer Power
Differentiate
\[ f(x)=x^5. \]Bring down the exponent \(5\), then reduce the exponent by \(1\):
\[ f'(x)=5x^{5-1}. \]Therefore,
\[ \boxed{f'(x)=5x^4}. \]Power Rule Example: A First Power
Since
\[ x=x^1, \]the power rule gives
\[ \frac{d}{dx}[x] =1x^{1-1} =x^0 =1. \]Thus,
\[ \boxed{\frac{d}{dx}[x]=1}. \]Power Rule Example: The Zero Power
For \(x\ne0\),
\[ x^0=1. \]Applying the power rule formally gives
\[ \frac{d}{dx}[x^0]=0x^{-1}=0, \]which agrees with the constant rule:
\[ \boxed{\frac{d}{dx}[1]=0}. \]Power Rule Example: A Negative Exponent
Rewrite the reciprocal expression
\[ f(x)=\frac{1}{x^3} \]as
\[ f(x)=x^{-3}. \]Apply the power rule:
\[ f'(x)=-3x^{-4}. \]In positive-exponent form,
\[ \boxed{f'(x)=-\frac{3}{x^4}},\qquad x\ne0. \]Power Rule Example: A Fractional Exponent
Consider
\[ f(x)=x^{1/2}=\sqrt{x}. \]On inputs where the derivative is defined, the power rule gives
\[ f'(x)=\frac12x^{-1/2}. \]Therefore,
\[ \boxed{f'(x)=\frac{1}{2\sqrt{x}}},\qquad x>0. \]Notice that the original function is defined at \(x=0\), but the displayed derivative formula is not. A derivative can have a more restricted domain than the original function.
Rewrite Radicals as Powers
The power rule becomes easier to use when radicals are rewritten with rational exponents.
For example,
\[ \sqrt[3]{x^2}=x^{2/3}. \]Therefore,
\[ \frac{d}{dx}[x^{2/3}] =\frac23x^{-1/3}. \]Equivalently,
\[ \boxed{\frac{d}{dx}[\sqrt[3]{x^2}]=\frac{2}{3\sqrt[3]{x}}},\qquad x\ne0. \]Why the Power Rule Is Plausible
For positive integer powers, the pattern can be seen from the limit definition. For example, let
\[ f(x)=x^3. \]Then
\[ \frac{f(x+h)-f(x)}{h} =\frac{(x+h)^3-x^3}{h}. \]Expand:
\[ (x+h)^3=x^3+3x^2h+3xh^2+h^3. \]Therefore,
\[ \frac{(x+h)^3-x^3}{h} =\frac{3x^2h+3xh^2+h^3}{h} =3x^2+3xh+h^2. \]Taking the limit as \(h\to0\) gives
\[ \boxed{f'(x)=3x^2}. \]The general power rule packages this recurring algebraic pattern into one reusable rule.
The Constant-Multiple Rule
A constant coefficient scales the derivative by the same amount.
If \(c\) is constant, then
\[ \boxed{\frac{d}{dx}[c f(x)]=c f'(x)}. \]The constant can be carried outside the differentiation process.
Constant-Multiple Example
Differentiate
\[ f(x)=7x^4. \]Keep the coefficient \(7\) and differentiate \(x^4\):
\[ f'(x)=7(4x^3). \]Thus,
\[ \boxed{f'(x)=28x^3}. \]Why the Constant-Multiple Rule Works
Let
\[ F(x)=c f(x). \]Then
\[ F'(x) =\lim_{h\to0}\frac{cf(x+h)-cf(x)}{h}. \]Factor out \(c\):
\[ F'(x) =c\lim_{h\to0}\frac{f(x+h)-f(x)}{h}. \]Therefore,
\[ \boxed{F'(x)=cf'(x)}. \]The Sum Rule
A sum can be differentiated term by term.
If \(f\) and \(g\) are differentiable, then
\[ \boxed{\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)}. \]Differentiation distributes across addition.
The Difference Rule
A difference can also be differentiated term by term:
\[ \boxed{\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)}. \]Differentiation preserves the subtraction between the two derivatives.
Why the Sum Rule Works
Let
\[ F(x)=f(x)+g(x). \]Its difference quotient is
\[ \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}. \]Rearrange the numerator:
\[ \frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}. \]Taking limits gives
\[ \boxed{F'(x)=f'(x)+g'(x)}. \]The difference rule follows by combining the sum rule with the constant-multiple rule using the constant \(-1\).
Term-by-Term Differentiation
The constant-multiple, sum, difference, and power rules work together to differentiate polynomial expressions one term at a time.
For example, if
\[ f(x)=4x^5-3x^2+7x-9, \]then
\[ f'(x) =4(5x^4)-3(2x)+7(1)-0. \]Therefore,
\[ \boxed{f'(x)=20x^4-6x+7}. \]A Practical Polynomial Shortcut
For a polynomial, you can usually work directly term by term:
\[ a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0. \]Differentiate each power term and drop the constant term:
\[ \boxed{ na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1 }. \]This is not a new rule. It is the repeated use of the rules already established.
Example: Differentiate a Polynomial
Differentiate
\[ y=6x^6-5x^4+2x^3-8x+11. \]Differentiate each term:
\[ \frac{dy}{dx} =36x^5-20x^3+6x^2-8. \]Thus,
\[ \boxed{y'=36x^5-20x^3+6x^2-8}. \]Example: Negative Powers
Differentiate
\[ f(x)=5x^{-2}-3x^{-5}. \]Apply the power rule term by term:
\[ f'(x)=5(-2)x^{-3}-3(-5)x^{-6}. \]Simplify:
\[ \boxed{f'(x)=-10x^{-3}+15x^{-6}}. \]In positive-exponent form,
\[ \boxed{f'(x)=-\frac{10}{x^3}+\frac{15}{x^6}},\qquad x\ne0. \]Example: Fractional Powers
Differentiate
\[ f(x)=4x^{3/2}-6x^{1/2}. \]Apply the power rule:
\[ f'(x) =4\left(\frac32\right)x^{1/2} -6\left(\frac12\right)x^{-1/2}. \]Therefore,
\[ \boxed{f'(x)=6\sqrt{x}-\frac{3}{\sqrt{x}}},\qquad x>0. \]Rewrite Before Differentiating
Algebraic rewriting can make a derivative much easier to see.
Suppose
\[ f(x)=\frac{4}{x^2}+3\sqrt{x}. \]Rewrite using powers:
\[ f(x)=4x^{-2}+3x^{1/2}. \]Then differentiate:
\[ f'(x)=-8x^{-3}+\frac32x^{-1/2}. \]Equivalently,
\[ \boxed{f'(x)=-\frac{8}{x^3}+\frac{3}{2\sqrt{x}}}. \]Simplify Before Using a More Complicated Rule
When an expression can be simplified into a sum of powers, do that before reaching for a more advanced differentiation rule.
For example,
\[ f(x)=\frac{x^4+3x^2}{x^2},\qquad x\ne0, \]simplifies to
\[ f(x)=x^2+3. \]Therefore,
\[ \boxed{f'(x)=2x},\qquad x\ne0. \]The original domain restriction still matters even after the algebraic expression becomes simpler.
Derivative Rules Preserve Linearity
The constant-multiple and sum rules together show that differentiation is a linear operation. For constants \(a\) and \(b\),
\[ \boxed{ \frac{d}{dx}[af(x)+bg(x)] =af'(x)+bg'(x) }. \]This single statement packages constant multiples, sums, and differences into one structural rule.
Example: A Linear Combination of Functions
Suppose
\[ F(x)=3f(x)-5g(x). \]If \(f\) and \(g\) are differentiable, then
\[ \boxed{F'(x)=3f'(x)-5g'(x)}. \]We do not need formulas for \(f\) and \(g\) to apply this structural rule.
Evaluate After Differentiating
When asked for a derivative at one input, it is usually efficient to find the derivative function first and substitute afterward.
Let
\[ f(x)=2x^4-3x^2+1. \]First differentiate:
\[ f'(x)=8x^3-6x. \]At \(x=2\),
\[ f'(2)=8(2)^3-6(2)=64-12=52. \]Therefore,
\[ \boxed{f'(2)=52}. \]Derivative Values Have Units
A derivative is a rate, so its units are output units divided by input units.
If position \(s(t)\) is measured in meters and time \(t\) in seconds, then
\[ s'(t) \]has units of meters per second.
The symbolic rules make differentiation fast, but the resulting derivative should still be interpreted as a rate of change.
Example: Position and Velocity
Suppose an object's position is
\[ s(t)=4t^3-6t^2+2t+5, \]where \(s\) is measured in meters and \(t\) in seconds.
Differentiate term by term:
\[ s'(t)=12t^2-12t+2. \]The derivative is the velocity function:
\[ \boxed{v(t)=12t^2-12t+2}. \]Its units are meters per second.
Example: Marginal Change
Suppose a modeled cost function is
\[ C(q)=0.02q^3+4q+500, \]where \(q\) is a production quantity.
Its derivative is
\[ C'(q)=0.06q^2+4. \]The derivative describes how the modeled cost is changing locally as production changes. In applications, this type of derivative is often interpreted as marginal cost.
Rules Do Not Replace Domain Awareness
Differentiation rules operate on formulas, but formulas live on domains. Always keep track of where the original function is defined and where the resulting derivative exists.
For example,
\[ f(x)=\sqrt{x} \]is defined for \(x\ge0\), while
\[ f'(x)=\frac{1}{2\sqrt{x}} \]is defined for \(x>0\).
A Derivative Formula Can Have a Smaller Domain
Differentiability is a stronger condition than merely being defined. A function may exist at an input where its derivative does not.
The function
\[ f(x)=x^{2/3} \]is defined at \(x=0\). For \(x\ne0\), the power rule gives
\[ f'(x)=\frac23x^{-1/3}. \]That derivative expression is undefined at \(x=0\). The original function therefore requires separate local analysis there rather than blind substitution into the rule-generated formula.
What These Rules Can Differentiate Directly
The rules in this lesson are enough to differentiate many expressions made from:
- constants;
- powers of the independent variable;
- constant multiples of those powers;
- sums of differentiable terms;
- differences of differentiable terms;
- polynomials;
- many radicals and reciprocals after rewriting them as powers.
What These Rules Do Not Yet Cover
Some expressions require additional structure. For example, the derivative of
\[ f(x)g(x), \]is not found by simply multiplying \(f'(x)\) and \(g'(x)\). Likewise, quotients and compositions require their own rules.
Those methods are developed in the next lesson: Product, Quotient, and Chain Rules.
Common Error: Forgetting to Reduce the Exponent
For
\[ f(x)=x^6, \]an incorrect result is
\[ f'(x)=6x^6. \]The exponent must decrease by \(1\):
\[ \boxed{f'(x)=6x^5}. \]Common Error: Dropping the Coefficient
For
\[ f(x)=8x^3, \]the coefficient \(8\) remains part of the calculation:
\[ f'(x)=8(3x^2). \]Therefore,
\[ \boxed{f'(x)=24x^2}. \]Common Error: Differentiating a Constant as Itself
A student may incorrectly write
\[ \frac{d}{dx}[12]=12. \]But a constant does not change with \(x\), so
\[ \boxed{\frac{d}{dx}[12]=0}. \]Common Error: Applying the Power Rule to a Sum as One Power
The power rule applies directly to an expression of the form \(x^n\). It does not justify treating a sum inside a power as though each term were an independent power.
For example,
\[ (x+1)^5 \]is a power of the entire expression \(x+1\), not merely a single power of \(x\). A systematic rule for differentiating compositions is developed in the chain-rule lesson.
Common Error: Multiplying Derivatives Termwise
The sum rule says
\[ (f+g)'=f'+g'. \]There is no parallel rule saying
\[ (fg)'=f'g'. \]Products require the product rule, which is introduced next.
Common Error: Ignoring a Negative Exponent
If
\[ f(x)=x^{-4}, \]then the exponent \(-4\) becomes the coefficient and is reduced by \(1\):
\[ f'(x)=-4x^{-5}. \]Therefore,
\[ \boxed{f'(x)=-\frac{4}{x^5}}. \]Common Error: Losing the Original Domain
Algebra may simplify a formula, but it does not retroactively add inputs excluded from the original function.
If
\[ f(x)=\frac{x^3}{x},\qquad x\ne0, \]then \(f(x)=x^2\) only on the original domain \(x\ne0\). Differentiating gives
\[ f'(x)=2x, \]still for
\[ \boxed{x\ne0}. \]A Reliable Procedure
- Identify the independent variable.
- Simplify the expression when useful.
- Rewrite radicals and reciprocals as powers when that makes the structure clearer.
- Separate sums and differences into terms.
- Keep constant coefficients attached to their terms.
- Apply the constant rule to constants.
- Apply the power rule to powers of the variable.
- Simplify the resulting derivative.
- Preserve relevant domain restrictions.
- Interpret units or rates when the problem has applied meaning.
Example: Full Workflow
Differentiate
\[ f(x)=\frac{3}{x^2}-4\sqrt{x}+5x^3-7. \]Rewrite the reciprocal and radical:
\[ f(x)=3x^{-2}-4x^{1/2}+5x^3-7. \]Differentiate term by term:
\[ f'(x) =3(-2)x^{-3} -4\left(\frac12\right)x^{-1/2} +15x^2. \]Simplify:
\[ \boxed{ f'(x)=-\frac{6}{x^3}-\frac{2}{\sqrt{x}}+15x^2 },\qquad x>0. \]Checking with a Limit at a Simple Point
Rules should agree with the limit definition. Consider
\[ f(x)=x^2. \]The power rule gives
\[ f'(x)=2x. \]At \(x=3\), this predicts
\[ f'(3)=6. \]From first principles,
\[ f'(3) =\lim_{h\to0}\frac{(3+h)^2-9}{h} =\lim_{h\to0}(6+h) =6. \]The efficient rule and the defining limit agree.
Practice 1: Constant Rule
Differentiate:
\[ f(x)=23. \]Practice 2: Power Rule
Differentiate:
\[ f(x)=x^9. \]Practice 3: Constant Multiple
Differentiate:
\[ f(x)=-6x^5. \]Practice 4: Polynomial
Differentiate:
\[ f(x)=3x^4-5x^3+2x-10. \]Practice 5: Negative Exponents
Differentiate:
\[ f(x)=2x^{-3}+7x^{-1}. \]Practice 6: Radical as a Power
Differentiate:
\[ f(x)=5\sqrt{x}. \]Practice 7: Mixed Powers
Differentiate:
\[ f(x)=4x^{3/2}-2x^{-2}+9. \]Practice 8: Evaluate a Derivative
Let
\[ f(x)=x^4-2x^2+3x. \]Find \(f'(2)\).
Practice 9: Rewrite First
Differentiate:
\[ f(x)=\frac{6}{x^4}+2\sqrt[3]{x}. \]Practice 10: Identify the Error
A student differentiates
\[ f(x)=5x^7-4 \]and writes
\[ f'(x)=35x^7-4. \]Identify and correct the errors.
Practice Answers
Practice 1
\[ \boxed{f'(x)=0}. \]Practice 2
\[ \boxed{f'(x)=9x^8}. \]Practice 3
\[ f'(x)=-6(5x^4), \]so
\[ \boxed{f'(x)=-30x^4}. \]Practice 4
\[ \boxed{f'(x)=12x^3-15x^2+2}. \]Practice 5
\[ f'(x)=-6x^{-4}-7x^{-2}. \]Equivalently,
\[ \boxed{f'(x)=-\frac{6}{x^4}-\frac{7}{x^2}},\qquad x\ne0. \]Practice 6
Rewrite \(\sqrt{x}=x^{1/2}\):
\[ f'(x)=5\left(\frac12\right)x^{-1/2}. \]Therefore,
\[ \boxed{f'(x)=\frac{5}{2\sqrt{x}}},\qquad x>0. \]Practice 7
\[ f'(x) =4\left(\frac32\right)x^{1/2} -2(-2)x^{-3}. \]Thus,
\[ \boxed{f'(x)=6x^{1/2}+4x^{-3}}. \]Practice 8
First differentiate:
\[ f'(x)=4x^3-4x+3. \]Then evaluate at \(x=2\):
\[ f'(2)=4(8)-8+3=27. \]Therefore,
\[ \boxed{f'(2)=27}. \]Practice 9
Rewrite:
\[ f(x)=6x^{-4}+2x^{1/3}. \]Differentiate:
\[ f'(x)=-24x^{-5}+\frac23x^{-2/3}. \]Therefore,
\[ \boxed{f'(x)=-\frac{24}{x^5}+\frac{2}{3x^{2/3}}},\qquad x\ne0. \]Practice 10
The exponent on \(x^7\) must decrease to \(6\), and the constant \(-4\) differentiates to \(0\). Therefore,
\[ \boxed{f'(x)=35x^6}. \]What to Remember
The foundational derivative rules are:
\[ \boxed{\frac{d}{dx}[c]=0} \] \[ \boxed{\frac{d}{dx}[x^n]=nx^{n-1}} \] \[ \boxed{\frac{d}{dx}[cf(x)]=cf'(x)} \] \[ \boxed{\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)} \] \[ \boxed{\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)}. \]Together, these rules make polynomial and power-function differentiation systematic. The most useful working habit is to simplify and rewrite the expression until its structure matches rules you already know.
Where We Go Next
The rules in this lesson handle constants, powers, scalar multiples, sums, and differences. More complicated functions often combine changing quantities through multiplication, division, or composition.
The next lesson, Product, Quotient, and Chain Rules, develops the derivative rules needed for those structures and shows how to recognize which rule a function requires.