Why Decompose a Composite Function?
A composite function is built by applying one function to the output of another. For example, in \(F(x)=\sqrt{4+3x}\), the expression \(4+3x\) is evaluated first, and the square root is applied to that result. This structure matters when finding a linear approximation: the rate of change of the whole function depends on the rates of change at each layer.
A linearization near a base point uses the function’s value and slope at that point. For a composite function, decomposing the expression makes both quantities easier to calculate correctly. The Chain Rule supplies the derivative, and the linearization formula then turns that derivative into a tangent-line model.
The Formula for a Composite Function
Suppose \(F=f\circ g\), meaning \(F(x)=f(g(x))\). Assume \(g\) is defined and differentiable near \(a\), and that \(f\) is defined and differentiable near \(g(a)\). These conditions ensure that the composition is defined near \(a\) and that the Chain Rule applies there. It gives
The function value is \(F(a)=f(g(a))\). Substituting this value and the Chain Rule derivative into the definition of linearization gives the composite formula.
The calculation has two stages. First, evaluate the inner function at the base point, \(g(a)\). Then evaluate the outer function and its derivative at that inner output. Finally, multiply the outer derivative by the inner derivative at \(a\). That product is the slope of the composite function at the base point.
Why the Formula Works
The formula follows directly from the definition of linearization and the Chain Rule. The following steps keep the function value and slope distinct, which is useful when the composition has several layers.
Since \(F(x)=f(g(x))\), we have \(F(a)=f(g(a))\).
The Chain Rule gives \(F'(x)=f'(g(x))g'(x)\), so \(F'(a)=f'(g(a))g'(a)\).
The definition \(L_{F,a}(x)=F(a)+F'(a)(x-a)\) yields \(f(g(a))+f'(g(a))g'(a)(x-a)\).
This also explains the role of a local approximation at each layer. Near \(a\), the inner function changes from \(g(a)\) at a rate \(g'(a)\). The outer function responds to a change in its input at rate \(f'(g(a))\). Multiplying the rates gives the rate of change from \(x\) through \(g(x)\) to \(f(g(x))\). The resulting linearization uses the actual composite value \(f(g(a))\) as its starting height.
For a nested expression with more than two layers, apply the same idea successively. For instance, if \(H(x)=r(f(g(x)))\), then its derivative at \(a\) is \(r'(f(g(a)))f'(g(a))g'(a)\). Each derivative is evaluated at the output of the layer immediately inside it. Keeping those inputs visible prevents the factors from being evaluated at the wrong point.
Worked Examples
Worked Example: Linearize a Square Root of a Linear Function
Find the linearization of \(F(x)=\sqrt{4+3x}\) at \(a=0\). Decompose the expression as \(g(x)=4+3x\) and \(f(u)=\sqrt{u}\), so \(F(x)=f(g(x))\).
First evaluate the inner function and its derivative at the base point:
The outer function and its derivative are \(f(u)=\sqrt{u}\) and \(f'(u)=\frac{1}{2\sqrt{u}}\). At the inner output \(u=4\),
Therefore \(F(0)=2\), and the Chain Rule gives
The linearization is
To check, substituting \(x=0\) into the linearization gives \(L_{F,0}(0)=2=F(0)\). Direct differentiation gives \(F'(x)=\frac{3}{2\sqrt{4+3x}}\), so \(F'(0)=\frac{3}{2\cdot 2}=\frac34\), matching the slope of the linearization.
Worked Example: Linearize an Exponential of a Trigonometric Function
Find the linearization of \(F(x)=e^{\sin x}\) at \(a=0\). Here \(g(x)=\sin x\) is the inner function and \(f(u)=e^u\) is the outer function. Trigonometric derivatives use radians.
Evaluate the inner function and its derivative:
For the outer function, \(f'(u)=e^u\), so \(f(0)=e^0=1\) and \(f'(0)=e^0=1\). Thus
Hence the linearization is
For the check, \(L_{F,0}(0)=1=F(0)\). Differentiating the original function by the Chain Rule gives \(F'(x)=e^{\sin x}\cos x\), and \(F'(0)=e^0\cos 0=1\), as required.
Worked Example: Linearize a Logarithm of a Quadratic
Find the linearization of \(F(x)=\ln(1+x^2)\) at \(a=1\). Set \(g(x)=1+x^2\) and \(f(u)=\ln u\). The logarithm requires a positive input; \(1+x^2>0\) for every real \(x\), so the composition is defined near the base point.
At \(a=1\), the inner value and derivative are
The outer function has derivative \(f'(u)=1/u\). Consequently, \(f(2)=\ln 2\) and \(f'(2)=1/2\). The composite value and derivative are
Therefore
As a check, \(L_{F,1}(1)=\ln 2=F(1)\). Differentiating directly through the composition gives \(F'(x)=\frac{1}{1+x^2}(2x)=\frac{2x}{1+x^2}\), and \(F'(1)=\frac{2}{2}=1\), which agrees with the linearization’s slope.
Worked Example: Organize a Three-Layer Composition
Find the linearization of \(H(x)=\sin(\sqrt{x})\) at \(a=4\). There are three layers: \(u(x)=\sqrt{x}\), \(v(u)=u\) as the input to the sine, and the outer function \(r(v)=\sin v\). Equivalently, use the two functions \(g(x)=\sqrt{x}\) and \(f(u)=\sin u\), with the square root as the inner layer.
The inner value and derivative are
The outer function satisfies \(f(2)=\sin 2\) and \(f'(2)=\cos 2\). Therefore
The linearization is
Substitution gives \(L_{H,4}(4)=\sin 2=H(4)\). Direct differentiation gives \(H'(x)=\cos(\sqrt{x})\frac{1}{2\sqrt{x}}\); evaluating at \(4\) gives \(H'(4)=\frac{\cos 2}{4}\), confirming the slope.
A Common Mistake: Dropping an Inner Derivative
In a composite function, the outer derivative alone is not usually the derivative of the whole expression. For \(F(x)=\sqrt{4+3x}\), using only the derivative of the square root at \(4\) would give the incorrect slope \(1/4\). That calculation treats the input to the square root as though it changed at rate \(1\), even though \(4+3x\) changes at rate \(3\).
The correction is to identify \(g'(a)\) and include it as a factor. A reliable check is to differentiate the original composite expression once and evaluate that derivative at the base point. If the slope in the proposed linearization does not match, a layer may have been missed or evaluated at the wrong input.
It is also useful to distinguish the linearization of a composition from the composition itself. The linearization of \(F\) is one linear function, \(F(a)+F'(a)(x-a)\). Simply replacing the inner function by a line inside the outer function generally produces a function that is not linear. The formula above directly supplies the value and slope of the whole composition, which are the two quantities needed for its linearization.
A Practical Calculation Pattern
When a formula contains several nested functions, record the layers from the inside out. At each layer, write down the input value at the base point and the derivative of that layer at that input. Then multiply the derivative factors in the Chain Rule and place the result into the linearization formula.
Use the specified value \(a\), and check that the composite is defined there and in a neighborhood where needed.
Name the inner function \(g\) and outer function \(f\), or list every layer for a deeper composition.
Find \(g(a)\), then the outer function value and derivative at that output.
Apply the Chain Rule with each derivative evaluated at the correct intermediate input.
Use \(F(a)+F'(a)(x-a)\), then confirm its value and slope at \(a\).
Check Your Understanding
For each question, identify the function layers before calculating. Check both the value and slope in any linearization you find.
- For \(F(x)=\sqrt{1+2x}\), identify \(f\) and \(g\), and state the conditions needed to use the Chain Rule at a base point \(a\).
- Find the linearization of \(F(x)=e^{2x}\) at \(a=0\).
- For \(F(x)=\ln(3+x^2)\), write \(F'(x)\) using the Chain Rule and identify where the outer derivative is evaluated.
- Explain why differentiating only the outer function is insufficient for \(F(x)=\sin(5x)\).
- For a three-layer composition \(r(f(g(x)))\), write its derivative at \(a\), indicating the input at which each factor is evaluated.
