Why the Input Transformation Matters
A function may be built from a familiar function by changing its input and multiplying its output. For example, \(3\sqrt{2x+7}\) uses a square root, but the square-root input is \(2x+7\), not \(x\). That change affects the slope, and therefore affects the local linear estimate.
The goal is to track both transformations carefully. If the input changes \(k\) times as fast as \(x\), the chain rule contributes a factor of \(k\) to the derivative. A multiplier outside the function contributes a factor of \(c\). Together, these factors determine the slope of the tangent-line model.
The Linearization Formula for a Composite Function
For a function \(h\) differentiable at \(a\), its linearization at \(a\) is the line that has the same value and slope as \(h\) at that point:
Now consider \(h(x)=c f(kx+b)\), where \(b\), \(c\), and \(k\) are constants. Define the inner input at the base point by \(u_0=ka+b\). If \(f\) is differentiable at \(u_0\), then the chain rule gives
Substituting these into the linearization formula gives a direct rule for this form of function.
The base point for the linearization is \(x=a\), in the original input variable. The corresponding base point for \(f\) is \(u_0=ka+b\). These are generally different numbers. The derivative \(f'\) must be evaluated at \(u_0\), while the final model estimates values of \(h\) at \(x\)-values near \(a\).
The formula also handles special cases. If \(c=0\), then \(h\) is the zero function and the linearization is zero. If \(k=0\), then the input to \(f\) is constant, so \(h\) is constant and the slope is zero. In every case, the required differentiability condition is that \(f\) be differentiable at \(ka+b\).
Why the Formula Works
The chain rule explains the slope, and the definition of linearization then determines the line. Here is the full calculation. Set \(u(x)=kx+b\), so \(h(x)=c f(u(x))\). Since \(u'(x)=k\), differentiating gives
At \(x=a\), the inner input is \(u(a)=ka+b\). Therefore \(h(a)=c f(ka+b)\) and \(h'(a)=ck f'(ka+b)\). A line through \((a,h(a))\) with slope \(h'(a)\) has equation \(h(a)+h'(a)(x-a)\), which is exactly the stated formula.
This also shows why the formula gives the appropriate local model: it matches both the function's height and its instantaneous rate of change at the base point. To verify the match directly, differentiate the linearization and evaluate it at \(a\):
The line therefore agrees with the function in value and slope at the chosen point. For inputs \(x\) near \(a\), the estimate is \(h(x)\approx L(x)\). The estimate is local: the formula specifies the tangent-line model, not a guarantee that the model is accurate far from \(a\).
Worked Example: A Scaled Square Root
Worked Example: A Scaled Square Root
Let \(h(x)=3\sqrt{2x+7}\). Find its linearization at \(a=1\), then use it to estimate \(h(1.02)\).
Write \(h(x)=c f(kx+b)\), where \(c=3\), \(f(u)=\sqrt{u}\), \(k=2\), and \(b=7\). The inner input at the base point is
Evaluate the function at the base point:
Since \(f'(u)=\frac{1}{2\sqrt{u}}\), the derivative of the composite at \(a=1\) is
Thus the linearization is
Substitute \(x=1.02\):
Check the model at the base point: \(L(1)=9=h(1)\). Its slope is \(L'(x)=1\), which matches \(h'(1)=1\). Substitution into the model gives \(L(1.02)=9.02\), the stated estimate.
Worked Example: A Shifted Logarithm
Worked Example: A Shifted Logarithm
Let \(h(x)=2\ln(5x-4)\). Find the linearization at \(a=1\), then estimate \(h(1.03)\). The logarithm requires \(5x-4>0\), or \(x>0.8\), so both the base point and the requested input are in the domain.
Here \(c=2\), \(f(u)=\ln u\), \(k=5\), and \(b=-4\). At the base point,
Since \(f'(u)=1/u\), the slope is
The linearization and estimate are therefore
Check the base value: \(L(1)=0=h(1)\). Differentiating the line gives \(L'(x)=10\), so \(L'(1)=10\). Directly differentiating \(h\) gives \(h'(x)=10/(5x-4)\), and hence \(h'(1)=10/1=10\). The estimate follows by substituting \(1.03\) into the checked linearization.
Worked Example: A Shifted and Reversed Cosine
Worked Example: A Shifted and Reversed Cosine
Let \(h(x)=-2\cos(3x+\pi/3)\). Find its linearization at \(a=0\), then estimate \(h(0.01)\).
In the composite form, \(c=-2\), \(f(u)=\cos u\), \(k=3\), and \(b=\pi/3\). The inner input at the base point is \(u_0=\pi/3\). Using \(\cos(\pi/3)=1/2\), the base value is
Because \(f'(u)=-\sin u\), the chain rule gives
Therefore,
At \(x=0.01\), the model gives
The decimal is rounded to six places. For the check, \(L(0)=-1=h(0)\), and \(L'(x)=3\sqrt{3}\). Direct differentiation gives \(h'(x)=6\sin(3x+\pi/3)\), so \(h'(0)=6(\sqrt{3}/2)=3\sqrt{3}\), as required.
Keeping the Factors Straight
The two constants \(c\) and \(k\) have different roles, even though both affect the slope. The factor \(c\) multiplies the function's value and derivative. The factor \(k\) comes from the rate of change of the inner input \(kx+b\); it multiplies the derivative through the chain rule. The shift \(b\) changes where \(f\) and \(f'\) are evaluated, but contributes no factor to the derivative because its derivative is zero.
A useful way to organize a calculation is to find three quantities before writing the line:
Compute \(u_0=ka+b\), the input to \(f\) when \(x=a\).
Compute \(h(a)=c f(u_0)\), keeping the outside multiplier \(c\).
Compute \(h'(a)=ck f'(u_0)\), then use \(L(x)=h(a)+h'(a)(x-a)\).
Common Mistake: Omitting the Inner Derivative
For \(h(x)=2\ln(5x-4)\) at \(a=1\), an incorrect calculation might use only the outside multiplier and write \(h'(1)=2f'(1)=2\). That omits the derivative of the inner function \(5x-4\), which is \(5\). It would produce the incorrect model \(L(x)=2(x-1)\) and the incorrect estimate \(L(1.03)=0.06\).
The correction is to include the chain-rule factor \(k=5\):
The correct line is \(L(x)=10(x-1)\), giving \(L(1.03)=10(0.03)=0.30\). Another related error is evaluating \(f'\) at \(a\) rather than at \(ka+b\). In this example they happen to be equal because both are \(1\), but that coincidence need not occur; the derivative of \(f\) belongs at the transformed input \(u_0\).
Using the Model Responsibly
Once the line is found, a nearby input can be written as \(x=a+\Delta x\). The estimate then takes the especially clear form
This makes the roles of the quantities explicit: start with the known base value and add the base-point slope times the small input change. The chain-rule formula supplies that slope for shifted and scaled composites. In applied problems, keep track of units as well: \(h'(a)\) has units of output per unit of input, so \(h'(a)\Delta x\) has units of output and can be added to \(h(a)\).
A correct formula does not by itself ensure a useful estimate. The input should be near the base point, and the function should be defined around the points being considered. The calculation here constructs the local model; judging whether it is sufficiently accurate for a particular purpose requires attention to the size of the change and the function's behavior nearby.
Check Your Understanding
For each question, identify the inner input at the base point before calculating the slope.
- For \(h(x)=4\sqrt{3x+1}\), find the linearization at \(x=1\).
- For \(h(x)=3e^{2x-1}\), identify the inner base input at \(x=1\) and calculate the linearization slope there.
- For \(h(x)=\ln(4x+1)\), find the linearization at \(x=0\), then use it to estimate \(h(0.02)\).
- In \(h(x)=c f(kx+b)\), explain why the derivative formula includes \(k\), and state where \(f'\) is evaluated for a linearization at \(x=a\).
- For \(h(x)=-\cos(2x)\), find the value and slope of its linearization at \(x=0\), and write the line.
