Concept understanding — Linear Approximation and Differentials
A nonlinear function is generally hard to evaluate exactly, but near any one point its graph looks almost like a straight line -- the tangent line at that point. This is the idea behind linear approximation.
Definition (Linear Approximation). Let f:(a,b)→R be differentiable and x0∈(a,b). The linear approximation L of f at x0 is
L(x)=f(x0)+f′(x0)(x−x0),∀x∈(a,b).
This is exactly the equation of the tangent line to y=f(x) at (x0,f(x0)). Because f is differentiable, f(x0+Δx)≈L(x0+Δx)=f(x0)+f′(x0)Δx when Δx is small, and the errorf(x)−L(x) shrinks to 0 faster than x→x0 (it is o(x−x0), a consequence of the definition of the derivative as a limit).
Tip
To linearly approximate an "ugly" value like 327.2 or (123)2/3: pick a nearby point x0 where the function and its derivative are easy to evaluate exactly (a perfect cube, a perfect fourth power, a multiple of 10, ...), compute f(x0) and f′(x0), and plug into L(x0+Δx)=f(x0)+f′(x0)Δx.
The differential. Writing Δx=dx, the tangent-line increment is called the differential of f:
df=f′(x)dx,equivalently df=f′(x)Δx.
Geometrically, Δf=f(x+dx)−f(x) is the actual rise along the curve, while df=f′(x)dx is the rise along the tangent line; for small dx, Δf≈df, but the two are generally not equal (only exactly equal when f is itself linear, f(x)=mx+c). df is a function of two independent quantities, x and dx -- not of x alone the way the derivative is.
Differentials of standard functions follow directly from the derivative rules, e.g. d(xn)=nxn−1dx, d(sinx)=cosxdx, d(ex)=exdx, d(logx)=x1dx. The algebraic properties of differentials mirror differentiation exactly:
and the differential is dF=∂x∂Fdx+∂y∂Fdy, with dx=Δx,dy=Δy. Geometrically this is the tangent plane to z=F(x,y) at (x0,y0) -- exactly as the one-variable linear approximation was a tangent line. The same pattern extends to three variables:
A linear approximation problem always has three ingredients: the base point (where the function is easy), the function value there, and the derivative(s) there. Get all three right and the rest is substitution.
Take f(x)=3x at the nearby perfect cube x0=27, where f and f′ are exact.
f(27)=3,f′(x)=31x−2/3⇒f′(27)=271.
L(x)=3+271(x−27); with Δx=0.2: 327.2≈3+270.2.
✓Final answer
L(x)=3+271(x−27), and 327.2≈3+1351≈3.0074.
We build the linear approximation L(x)=f(x0)+f′(x0)(x−x0) of f(x)=3x at the perfect cube x0=27 (where f,f′ are exact), then evaluate L at x=27.2.
Step 1. Set up f and locate a convenient base point.f(x)=x1/3. Since 27=33 is a perfect cube near 27.2, take x0=27, so Δx=27.2−27=0.2.
Step 2. Compute f(x0) and f′(x0).f(27)=271/3=3. f′(x)=31x−2/3, so f′(27)=31⋅272/31=31⋅91=271 (using 272/3=(271/3)2=32=9).
Step 3. Write the linear approximation. By Definition 8.1, L(x)=f(27)+f′(27)(x−27)=3+271(x−27).
Step 4. Evaluate at x=27.2.L(27.2)=3+271(0.2)=3+270.2=3+1351.
Step 5. Simplify.1351≈0.007407, so 327.2≈3.0074 (the true value is 327.2≈3.00739, an excellent match).
✓Final answer
L(x)=3+271(x−27), and 327.2≈3+1351≈3.0074.
Choosing x0=27.2 itself instead of a nearby point where f,f′ are exactly computable
Errors simplifying 272/3 — forgetting it equals (271/3)2=9, not 272