Linear Approximation and Differentials

$$f(x)=x^6-7x^5+4x^4-x^3+3x^2+2x-8$$

The recording uses this polynomial to show why a simpler local model can be useful.

$$f(0)=-8,\qquad f'(0)=2$$

Differentiate first, then evaluate at the base point.

$$L_0(x)=-8+2x,\qquad f(0.1)\approx L_0(0.1)=-7.8$$

This estimate avoids substituting 0.1 into all seven terms. It is an approximation, not an exact polynomial value.

$$f(1)=-6,\qquad f'(1)=6-35+16-3+6+2=-8$$

Use the original polynomial from the preceding example.

$$L_1(x)=-6-8(x-1)=-8x+2$$

This line is anchored at x=1. It differs from the tangent line at zero.

$$f(0.9)\approx L_1(0.9)=-8(0.9)+2=-5.2$$

Use the arithmetic correction made during the recording. The actual polynomial value is −5.476589. This line is useful near one; the line at zero models a different neighborhood.

$$f(x)=\sin x,\qquad f(0)=0,\qquad f'(0)=\cos0=1$$

Use radians.

$$L(x)=x,\qquad\sin(0.2)\approx0.2$$

The calculator comparison in the lesson gives about 0.198669 for the actual value. The approximation is slightly high.

$$f(x)=\ln x,\qquad f(1)=0,\qquad f'(1)=1$$

The real logarithm requires x>0. One is a convenient nearby base point.

$$L(x)=x-1,\qquad\ln(1.2)\approx0.2$$

The actual value is about 0.182322. Since ln x is concave down, its tangent line gives an overestimate.

$$f(x)=\sqrt{x},\qquad f(9)=3,\qquad f'(9)=\frac1{2\sqrt9}=\frac16$$

The derivative formula requires x>0.

$$L(x)=3+\frac16(x-9)$$

The base point is nine, not 9.3.

$$\sqrt{9.3}\approx3+\frac{0.3}{6}=3.05$$

The actual value is about 3.049590, so the estimate is slightly high.

$$y=\sin x,\qquad\frac{dy}{dx}=\cos x$$

The derivative describes the slope.

$$dy=\cos x\,dx$$

The differential describes the tangent-line change for a chosen input increment dx. It is generally an estimate for the corresponding actual change in sine.

$$dx=\frac{dy}{\cos x}\qquad(\cos x\ne0)$$

Solving this differential relation for dx requires a nonzero slope. At a point where cosine is zero, this division cannot be used.