Local Linear Approximation and Differentiability

1Introduction

This lecture explains differentiation not only as the slope of a tangent line but also as a method for approximating the increment of a function near a base point by a linear map.

2Geometric Interpretation

The graph of a differentiable function can be approximated near a base point by its tangent line. This tangent-line approximation is called a local linear approximation.

At a base point $$a$$a, the approximation is

Setting $$h=x-a$$h=x-a, the map $$L\left(h\right)=f^{\prime }\left(a\right)h$$L(h)=f'(a)h acting on the increment is a linear map from $$\mathbb{R}$$\mathbb{R} to $$\mathbb{R}$$\mathbb{R}. In contrast, $$f\left(a\right)+L\left(h\right)$$f(a)+L(h) is an affine map because it includes a constant term. The word linear in local linear approximation refers to the linearity of $$L$$L as a function of the increment.

3Precise Characterization

Let $$D\subseteq \mathbb{R}$$D\subseteq\mathbb{R}, $$f:D\to \mathbb{R}$$f:D\to\mathbb{R}, and $$a\in \mathrm{int}D$$a\in\operatorname{int}D. The function $$f$$f is differentiable at $$a$$a if and only if there are a real number $$c$$c and a remainder $$r\left(h\right)$$r(h) such that

with

This condition is also written as $$r\left(h\right)=o\left(h\right)$$r(h)=o(h). It means that $$\left|r\right(h\left)\right|/\left|h\right|\to 0$$|r(h)|/|h|\to0, so the absolute error is negligible relative to $$\left|h\right|$$|h|. Indeed, for $$h\ne 0$$h\ne0, division by $$h$$h gives

so the difference quotient tends to $$c$$c and hence $$c=f^{\prime }\left(a\right)$$c=f'(a). Conversely, if $$f^{\prime }\left(a\right)$$f'(a) exists, defining $$r\left(h\right)=f(a+h)-f\left(a\right)-f^{\prime }\left(a\right)h$$r(h)=f(a+h)-f(a)-f'(a)h gives $$r\left(h\right)/h\to 0$$r(h)/h\to0.

4Example

As an example, approximate $$\sqrt{4.1}$$\sqrt{4.1}. Let $$f:(0,\mathrm{\infty })\to \mathbb{R}$$f:(0,\infty)\to\mathbb{R}, $$f\left(x\right)=\sqrt{x}$$f(x)=\sqrt{x}, and $$a=4$$a=4. Rationalizing the difference quotient gives

so $$f^{\prime }\left(4\right)=1/4$$f'(4)=1/4. Therefore,

and substitution of $$x=4.1$$x=4.1 yields

The actual value is $$\sqrt{4.1}\approx 2.02485$$\sqrt{4.1}\approx2.02485, confirming that the error of the first-order approximation is small near the base point.

5Related Material

6Exercises