The Central Idea
Suppose \(f\) is differentiable at \(x=a\). Near \(a\), the graph of \(f\) is well approximated by its tangent line.
The tangent line passes through \((a,f(a))\) and has slope \(f'(a)\). Therefore,
\[ \boxed{ L(x)=f(a)+f'(a)(x-a) }. \]The function \(L(x)\) is the linearization of \(f\) at \(x=a\).
Why Linearization Works
The derivative measures local rate of change. If \(x\) changes only a small amount from \(a\), the output change is approximately the local slope multiplied by the input change:
\[ f(x)-f(a) \approx f'(a)(x-a). \]Rearranging gives
\[ \boxed{ f(x)\approx f(a)+f'(a)(x-a) }. \]The approximation is usually strongest when \(x\) is close to \(a\).
Linearization Is the Tangent Line
The point-slope equation of the tangent line at \(x=a\) is
\[ y-f(a)=f'(a)(x-a). \]Solving for \(y\) gives
\[ y=f(a)+f'(a)(x-a). \]This is exactly the linearization formula. Linearization is therefore the tangent line used as a local model.
The Three Pieces of a Linearization
In
\[ L(x)=f(a)+f'(a)(x-a), \]- \(f(a)\) gives the starting output value;
- \(f'(a)\) gives the local slope or sensitivity;
- \(x-a\) gives the displacement from the base point.
The formula combines position, local direction, and input displacement.
Example: Linearizing \(x^2\)
Let
\[ f(x)=x^2 \]and linearize at \(a=3\).
\[ f(3)=9. \]Since
\[ f'(x)=2x, \]we have
\[ f'(3)=6. \]Therefore,
\[ L(x)=9+6(x-3). \]Equivalently,
\[ \boxed{ L(x)=6x-9 }. \]Using the Linearization
Use the linearization of \(x^2\) at \(x=3\) to estimate \(3.02^2\).
Since
\[ L(x)=9+6(x-3), \]we obtain
\[ L(3.02) = 9+6(0.02) = 9.12. \]Therefore,
\[ \boxed{ 3.02^2\approx9.12 }. \]The exact value is \(9.1204\), so the first-order estimate is very close.
Choose a Convenient Base Point
A good base point should satisfy two conditions:
- it should be close to the target input;
- the function and derivative should be easy to evaluate there.
To estimate
\[ \sqrt{16.4}, \]a natural base point is \(a=16\) because \(\sqrt{16}=4\).
Example: Linearizing a Square Root
Let
\[ f(x)=\sqrt{x} \]and choose \(a=16\).
\[ f(16)=4. \]Differentiate:
\[ f'(x)=\frac{1}{2\sqrt{x}}. \]Then
\[ f'(16)=\frac18. \]Therefore,
\[ \boxed{ L(x)=4+\frac18(x-16) }. \]Estimating \(\sqrt{16.4}\)
Substitute \(x=16.4\):
\[ L(16.4) = 4+\frac18(0.4) = 4.05. \]Thus,
\[ \boxed{ \sqrt{16.4}\approx4.05 }. \]Linearization Is a First-Order Approximation
The value \(f(a)\) gives the starting level, and \(f'(a)\) gives the first-order change.
Linearization does not include the effects of
\[ f''(a),\qquad f^{(3)}(a),\qquad \ldots \]so it ignores higher-order bending information.
That is why a linearization is also called a first-order approximation.
The Small-Change Form
Suppose
\[ x=a+\Delta x. \]Then
\[ x-a=\Delta x. \]The linearization becomes
\[ f(a+\Delta x) \approx f(a)+f'(a)\Delta x. \]Therefore,
\[ \boxed{ \Delta y\approx f'(a)\Delta x }. \]This is the basic small-change approximation.
What Is a Differential?
If
\[ y=f(x), \]the differential of \(x\) is written \(dx\), and the differential of \(y\) is defined by
\[ \boxed{ dy=f'(x)\,dx }. \]The quantity \(dy\) represents the change predicted by the local tangent line for an input change \(dx\).
Actual Change and Differential Change
If the input changes from \(x\) to \(x+\Delta x\), then the actual output change is
\[ \boxed{ \Delta y = f(x+\Delta x)-f(x) }. \]The differential change is
\[ \boxed{ dy=f'(x)\,dx }. \]When we choose \(dx=\Delta x\), then for sufficiently small changes,
\[ \boxed{ \Delta y\approx dy }. \]Do Not Confuse \(dy\) with \(\Delta y\)
The quantity \(\Delta y\) follows the actual curve.
The quantity \(dy\) follows the tangent line.
They are generally not identical. Their usefulness comes from the fact that they become close when the input change is small and the function is differentiable near the base point.
Example: Comparing \(dy\) and \(\Delta y\)
Let
\[ y=x^2 \]and let \(x\) change from \(3\) to \(3.02\).
Then
\[ dx=\Delta x=0.02. \]The differential is
\[ dy = 2x\,dx = 2(3)(0.02) = 0.12. \]The actual change is
\[ \Delta y = 3.02^2-3^2 = 9.1204-9 = 0.1204. \]Thus,
\[ \boxed{ dy=0.12, \qquad \Delta y=0.1204 }. \]The Differential Form of the Derivative
If
\[ \frac{dy}{dx}=f'(x), \]then differential notation writes the local relationship as
\[ \boxed{ dy=f'(x)\,dx }. \]This notation is extremely useful, but the derivative should not be treated as an ordinary fraction in every mathematical context.
Example: Differential of a Power
If
\[ y=x^5, \]then
\[ \frac{dy}{dx}=5x^4. \]Therefore,
\[ \boxed{ dy=5x^4\,dx }. \]Example: Estimating a Fifth Power
Estimate
\[ 2.01^5. \]Let
\[ f(x)=x^5 \]and choose \(a=2\).
\[ f(2)=32. \]Also,
\[ f'(2)=5(2)^4=80. \]Since \(2.01-2=0.01\),
\[ f(2.01) \approx 32+80(0.01) = 32.8. \]Thus,
\[ \boxed{ 2.01^5\approx32.8 }. \]Derivatives as Local Sensitivity
The derivative \(f'(a)\) tells us how strongly the output responds to small changes in the input near \(a\).
If \(|f'(a)|\) is large, a small input change can produce a relatively large output change.
If \(|f'(a)|\) is small, the same-size input change may have a smaller local effect.
Differentials as Sensitivity Calculations
The equation
\[ dy=f'(x)\,dx \]can be read as
\[ \boxed{ \text{output response} = \text{local sensitivity} \times \text{input perturbation} }. \]This interpretation appears throughout modeling, engineering, economics, numerical analysis, and scientific measurement.
Measurement Error
Suppose a measured quantity \(x\) has a small possible error \(dx\), and a derived quantity is
\[ y=f(x). \]The resulting output error can be estimated by
\[ \boxed{ dy\approx f'(x)\,dx }. \]For a magnitude estimate, use absolute values:
\[ \boxed{ |dy|\approx|f'(x)|\,|dx| }. \]Example: Error in Circle Area
Suppose the radius of a circle is measured as
\[ r=10\text{ cm} \]with possible error
\[ |dr|\le0.02\text{ cm}. \]Since
\[ A=\pi r^2, \]we have
\[ dA=2\pi r\,dr. \]Therefore,
\[ |dA| \approx 2\pi(10)(0.02) = 0.4\pi. \]So the area error is approximately
\[ \boxed{ |dA|\approx0.4\pi\text{ cm}^2 }. \]Relative Error
Absolute error is expressed in the units of the quantity.
Relative error compares the error with the size of the quantity itself:
\[ \boxed{ \frac{|dy|}{|y|} }. \]Relative error is unitless.
Percentage Error
Multiply relative error by \(100\%\):
\[ \boxed{ \text{percentage error} \approx \frac{|dy|}{|y|}\times100\% }. \]Percentage error is often easier to compare across quantities of different sizes.
Relative Error for Circle Area
For
\[ A=\pi r^2, \]we have
\[ dA=2\pi r\,dr. \]Divide by \(A\):
\[ \frac{dA}{A} = \frac{2\pi r\,dr}{\pi r^2}. \]Therefore,
\[ \boxed{ \frac{dA}{A} = 2\frac{dr}{r} }. \]A small relative radius error produces approximately twice that relative error in area.
Relative Error for Sphere Volume
For
\[ V=\frac43\pi r^3, \]we have
\[ dV=4\pi r^2\,dr. \]Divide by \(V\):
\[ \frac{dV}{V} = \frac{4\pi r^2\,dr}{\frac43\pi r^3}. \]Thus,
\[ \boxed{ \frac{dV}{V} = 3\frac{dr}{r} }. \]A small relative radius error is amplified by a factor of approximately \(3\) in the computed volume.
Power-Law Sensitivity
Suppose
\[ y=Cx^n \]where \(C\) is constant.
Then
\[ dy=Cnx^{n-1}\,dx. \]Divide by \(y=Cx^n\):
\[ \boxed{ \frac{dy}{y} = n\frac{dx}{x} }. \]The exponent directly controls first-order relative sensitivity.
Linearization Near Zero
Many useful approximations are centered at \(a=0\).
Then
\[ L(x)=f(0)+f'(0)x. \]This often produces compact formulas for small values of \(x\).
Example: \(\sqrt{1+x}\)
Let
\[ f(x)=\sqrt{1+x}. \]At \(x=0\),
\[ f(0)=1. \]Also,
\[ f'(x)=\frac{1}{2\sqrt{1+x}}, \qquad f'(0)=\frac12. \]Therefore,
\[ \boxed{ \sqrt{1+x} \approx 1+\frac{x}{2} } \]for sufficiently small \(x\).
Example: Reciprocal Approximation
Let
\[ f(x)=\frac{1}{1+x}. \]Since
\[ f(0)=1 \]and
\[ f'(0)=-1, \]the linearization is
\[ \boxed{ \frac{1}{1+x} \approx 1-x } \]for small \(x\).
Example: Exponential Approximation
For
\[ f(x)=e^x, \]we have
\[ f(0)=1 \]and
\[ f'(0)=1. \]Therefore,
\[ \boxed{ e^x\approx1+x } \]for small \(x\).
Example: Small-Angle Approximation
Let
\[ f(x)=\sin x. \]At \(x=0\),
\[ f(0)=0, \qquad f'(0)=1. \]Therefore,
\[ \boxed{ \sin x\approx x } \]for small \(x\) measured in radians.
Another Small-Angle Approximation
For
\[ f(x)=\tan x, \]we have
\[ f(0)=0 \]and
\[ f'(0)=\sec^2(0)=1. \]Thus,
\[ \boxed{ \tan x\approx x } \]for small radian values of \(x\).
Linearization of Composite Functions
The formula
\[ L(x)=f(a)+f'(a)(x-a) \]works for composite functions too.
The main additional task is computing \(f'(a)\) correctly, often using the chain rule.
Example: Composite Linearization
Let
\[ f(x)=\sqrt{x^2+9} \]and linearize at \(a=4\).
First,
\[ f(4)=5. \]Differentiate:
\[ f'(x) = \frac{x}{\sqrt{x^2+9}}. \]Then
\[ f'(4)=\frac45. \]Therefore,
\[ \boxed{ L(x) = 5+\frac45(x-4) }. \]Approximation Error
Define the linearization error by
\[ \boxed{ E(x)=f(x)-L(x) }. \]At the base point,
\[ E(a)=0. \]Because the tangent line also matches the slope,
\[ E'(a)=0. \]The remaining error comes from higher-order behavior of the function.
Why the Error Usually Grows Farther Away
A tangent line matches a function's position and slope at one point.
As we move farther from that point, curvature can cause the graph and tangent line to separate.
Therefore, a linearization should be viewed as a local model rather than a global replacement for the function.
Curvature Helps Explain Approximation Quality
The second derivative describes how the slope changes.
If \(|f''(x)|\) is small near the base point, the tangent line may remain a good approximation over a larger neighborhood.
If \(|f''(x)|\) is large, the graph may bend away from its tangent line more rapidly.
Concavity will be studied systematically in the next lesson.
Overestimates and Underestimates
Linearization can lie above or below the actual function.
When a function is locally concave up, the tangent line typically lies below the curve near the point of tangency.
When a function is locally concave down, the tangent line typically lies above the curve near the point of tangency.
Example: Why the Square-Root Estimate Is Slightly High
For
\[ f(x)=\sqrt{x}, \]the second derivative is
\[ f''(x) = -\frac{1}{4x^{3/2}}. \]For \(x>0\),
\[ f''(x)<0. \]The graph is concave down, so its tangent line lies slightly above the curve near the point of tangency. This explains why
\[ \sqrt{16.4}\approx4.05 \]is a slight overestimate.
Differentials and Units
Suppose \(x\) is measured in meters and \(y=f(x)\) is measured in square meters.
Then \(f'(x)\) has units of square meters per meter, while \(dx\) has units of meters.
Their product
\[ dy=f'(x)\,dx \]correctly has units of square meters.
Why Linearization Matters Beyond Hand Estimation
Linearization is a foundational modeling technique because nonlinear systems are often easier to analyze locally through linear approximations.
It supports:
- numerical approximation;
- measurement-error analysis;
- sensitivity analysis;
- small-perturbation modeling;
- local stability reasoning;
- higher-dimensional linear models.
A Reliable Procedure for Linearization
- Identify the function \(f(x)\).
- Choose a convenient base point \(a\) near the target input.
- Compute \(f(a)\).
- Compute \(f'(x)\).
- Evaluate \(f'(a)\).
- Write \(L(x)=f(a)+f'(a)(x-a)\).
- Substitute the desired nearby input into \(L(x)\).
- Use \(\approx\) unless the result is known to be exact.
A Reliable Procedure for Differentials
- Write the relationship \(y=f(x)\).
- Differentiate to find \(f'(x)\).
- Write \(dy=f'(x)\,dx\).
- Identify the base value of \(x\).
- Identify the small change or uncertainty \(dx\).
- Compute \(dy\).
- Interpret \(dy\) as an approximate output change or propagated error.
- Include units and relative-error information when appropriate.
Common Error: Choosing a Poor Base Point
To estimate
\[ \sqrt{24.9}, \]the base point \(a=25\) is especially useful because it is close to the target and
\[ \sqrt{25}=5. \]A base point should be both nearby and convenient.
Common Error: Forgetting \(x-a\)
The correct formula is
\[ L(x)=f(a)+f'(a)(x-a), \]not
\[ f(a)+f'(a)x. \]The quantity \(x-a\) measures displacement from the point of tangency.
Common Error: Using \(f'(x)\) Instead of \(f'(a)\)
The tangent line at \(x=a\) has one fixed slope:
\[ f'(a). \]The expression
\[ f(a)+f'(x)(x-a) \]generally does not describe a line because \(f'(x)\) may vary with \(x\).
Common Error: Treating an Approximation as Exact
Linearization usually gives
\[ f(x)\approx L(x), \]not
\[ f(x)=L(x). \]The equality is exact at the base point, but generally not elsewhere.
Common Error: Assuming \(dy=\Delta y\)
The actual change is
\[ \Delta y=f(x+\Delta x)-f(x). \]The differential is
\[ dy=f'(x)\,dx. \]The useful relation is
\[ \Delta y\approx dy \]when the change is sufficiently small.
Common Error: Using a Large Change as Though It Were Small
Linearization is local.
If the target input is far from the base point, the tangent line may no longer approximate the function well.
Choose a closer base point or use a more appropriate approximation method.
Common Error: Ignoring Domain Restrictions
A linear model does not erase the domain of the original function.
For example,
\[ f(x)=\sqrt{x} \]still requires
\[ x\ge0. \]The tangent line may be defined at values where the original function is not, but that does not make those values valid inputs for \(f\).
Practice 1: Basic Linearization
Find the linearization of
\[ f(x)=x^3 \]at \(a=2\).
Practice 2: Square-Root Estimate
Use linearization to estimate
\[ \sqrt{25.5}. \]Practice 3: Reciprocal Estimate
Use the linearization of
\[ f(x)=\frac1x \]at \(a=4\) to estimate
\[ \frac1{4.1}. \]Practice 4: Differential Change
If
\[ y=x^4, \]estimate the change in \(y\) when \(x\) changes from \(2\) to \(2.03\).
Practice 5: Actual Change versus Differential
For
\[ y=x^2, \]let \(x\) change from \(5\) to \(5.01\). Find both \(dy\) and \(\Delta y\).
Practice 6: Measurement Error
The radius of a circle is measured as \(12\) cm with possible error \(0.03\) cm. Use differentials to estimate the resulting maximum area error.
Practice 7: Relative Error
If the radius of a sphere has relative measurement error of approximately \(0.4\%\), estimate the corresponding relative error in the computed volume.
Practice 8: Small-Angle Approximation
Use
\[ \sin x\approx x \]to estimate \(\sin(0.03)\), where the angle is in radians.
Practice 9: Composite Linearization
Find the linearization of
\[ f(x)=\sqrt{x^2+5} \]at \(a=2\).
Practice 10: Identify the Error
A student linearizes \(f(x)\) at \(x=a\) using
\[ L(x)=f(a)+f'(x)(x-a). \]Explain the error and give the correct formula.
Practice Answers
Practice 1
For
\[ f(x)=x^3, \]we have
\[ f(2)=8, \qquad f'(x)=3x^2, \qquad f'(2)=12. \]Therefore,
\[ \boxed{ L(x)=8+12(x-2) }. \]Practice 2
Let \(f(x)=\sqrt{x}\) and choose \(a=25\).
\[ f(25)=5, \qquad f'(25)=\frac1{10}. \]Therefore,
\[ L(x)=5+\frac1{10}(x-25). \]At \(x=25.5\),
\[ \boxed{ \sqrt{25.5}\approx5.05 }. \]Practice 3
For
\[ f(x)=\frac1x, \]at \(a=4\),
\[ f(4)=\frac14, \qquad f'(4)=-\frac1{16}. \]Thus,
\[ L(x) = \frac14-\frac1{16}(x-4). \]At \(x=4.1\),
\[ \boxed{ \frac1{4.1}\approx0.24375 }. \]Practice 4
Since
\[ y=x^4, \]we have
\[ dy=4x^3\,dx. \]At \(x=2\) and \(dx=0.03\),
\[ dy = 4(2)^3(0.03) = 0.96. \]Therefore,
\[ \boxed{ \Delta y\approx0.96 }. \]Practice 5
Here
\[ x=5, \qquad dx=\Delta x=0.01. \]Therefore,
\[ dy = 2(5)(0.01) = 0.10. \]The actual change is
\[ \Delta y = 5.01^2-5^2 = 0.1001. \]Thus,
\[ \boxed{ dy=0.10, \qquad \Delta y=0.1001 }. \]Practice 6
For
\[ A=\pi r^2, \]we have
\[ dA=2\pi r\,dr. \]With \(r=12\) and \(|dr|=0.03\),
\[ |dA| \approx 2\pi(12)(0.03) = 0.72\pi. \]Therefore,
\[ \boxed{ |dA|\approx0.72\pi\text{ cm}^2 }. \]Practice 7
For a sphere,
\[ \frac{dV}{V} = 3\frac{dr}{r}. \]Therefore,
\[ 3(0.4\%)=1.2\%. \]So,
\[ \boxed{ \text{relative volume error}\approx1.2\% }. \]Practice 8
For a small radian angle,
\[ \sin x\approx x. \]Thus,
\[ \boxed{ \sin(0.03)\approx0.03 }. \]Practice 9
Let
\[ f(x)=\sqrt{x^2+5}. \]Then
\[ f(2)=3 \]and
\[ f'(x) = \frac{x}{\sqrt{x^2+5}}. \]Therefore,
\[ f'(2)=\frac23. \]The linearization is
\[ \boxed{ L(x) = 3+\frac23(x-2) }. \]Practice 10
A linearization at \(x=a\) uses the fixed tangent slope \(f'(a)\), not the variable slope \(f'(x)\).
The correct formula is
\[ \boxed{ L(x)=f(a)+f'(a)(x-a) }. \]What to Remember
Linearization replaces a differentiable function locally by its tangent line:
\[ \boxed{ L(x)=f(a)+f'(a)(x-a) }. \]Near \(x=a\),
\[ \boxed{ f(x)\approx L(x) }. \]Small changes satisfy
\[ \boxed{ \Delta y\approx f'(x)\Delta x }. \]Differential notation expresses the same local model as
\[ \boxed{ dy=f'(x)\,dx }. \]Keep these distinctions clear:
- \(\Delta y\) is the actual change;
- \(dy\) is the tangent-line change;
- linearization is most accurate near the base point;
- the derivative measures local sensitivity;
- differentials can estimate propagated measurement error;
- relative error compares error with the size of the quantity;
- curvature influences how quickly the approximation loses accuracy.
The reusable principle is
\[ \boxed{ \text{small output change} \approx \text{local derivative} \times \text{small input change} }. \]Where We Go Next
Linearization uses derivative information to build a local approximation. Derivatives can also describe the broader behavior of a function across an interval.
The next lesson, Critical Points, Monotonicity, and Concavity, uses first and second derivatives to determine where functions increase, decrease, flatten, bend upward, bend downward, and change qualitative behavior.