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.
(ii) f=x2+2x+3: Δf=f(−0.4)−f(−0.5)=0.11; df=(2x+2)(0.1)∣x=−0.5=0.1.
✓Final answer
(i) Δf=3.125,df=2 (ii) Δf=0.11,df=0.1 — in each case Δf≈df, with the gap coming from Δx not being infinitesimally small.
Compute the exact change Δf=f(x+Δx)−f(x) directly, and the approximate (tangent-line) change df=f′(x)dx, then compare.
Part (i): f(x)=x3−2x2;x=2,Δx=dx=0.5.
Step 1. Exact change Δf.f(2)=8−8=0. f(2.5)=15.625−2(6.25)=15.625−12.5=3.125. So Δf=f(2.5)−f(2)=3.125−0=3.125.
Step 2. Approximate change df.f(x)=x3−2x2⇒f′(x)=3x2−4x. At x=2: f′(2)=3(4)−4(2)=12−8=4. So df=f′(2)dx=4(0.5)=2.
Step 3. Compare.Δf=3.125 and df=2 — close but not equal, since Δx=0.5 is not infinitesimally small; the gap Δf−df=1.125 is the higher-order error the linear approximation ignores.
Part (ii): f(x)=x2+2x+3;x=−0.5,Δx=dx=0.1.
Step 1. Exact change Δf.f(−0.5)=0.25−1+3=2.25. f(−0.4)=0.16−0.8+3=2.36. So Δf=2.36−2.25=0.11.
Step 2. Approximate change df.f′(x)=2x+2, so f′(−0.5)=−1+2=1. df=f′(−0.5)dx=1(0.1)=0.1.
Step 3. Compare.Δf=0.11 and df=0.1 — again close, with a small gap of 0.01 from the finite (not infinitesimal) Δx.
✓Final answer
(i) Δf=3.125,df=2 (ii) Δf=0.11,df=0.1 — both cases confirm Δf≈df.
Direct comparison of exact increment Δf vs. differential df=f′(x)dx
Computing f′(2) incorrectly as 3x2−4x evaluated with a sign slip (writing −4x as +4x)
Evaluating Δf at the wrong endpoint, e.g. f(2)−f(2.5) instead of f(2.5)−f(2)