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Mathematics

Derivatives

A derivative gives the slope of a curve at one point. It comes from asking what happens to the slope between two points as they meet.
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From average change to change right now

The slope between two points is a secant slope. Make their horizontal gap smaller and the secant turns toward a tangent. The limit of those secant slopes is the derivative.

secant slope=f(a+h)−f(a)h,f′(a)=lim⁡h→0f(a+h)−f(a)h\text{secant slope}=\frac{f(a+h)-f(a)}{h},\qquad f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

For example, if position changes with time, this same idea turns an average velocity over an interval into velocity at an instant.

Shrink a secant into a tangent

Move the point and shrink the horizontal gap. Compare the two slopes.

Blue is f(x)=x2f(x)=x^2. Amber is the secant through two points; dashed green is the tangent at the selected point. The horizontal axis is x∈[−3,3]x\in[-3,3].
a=1a=1
h=0.316h=0.316
f(a+h)−f(a)h=2.316\frac{f(a+h)-f(a)}{h}=2.316
f′(a)=2a=2f'(a)=2a=2

As the gap approaches zero from either side, the secant slope approaches the tangent slope.

Why the square curve has this slope

Expand the numerator of the secant slope. Once the nonzero gap cancels, letting the gap approach zero is straightforward.

(a+h)2−a2h=2ah+h2h=2a+h(h≠0)\frac{(a+h)^2-a^2}{h}=\frac{2ah+h^2}{h}=2a+h\quad(h\ne0)
lim⁡h→0(2a+h)=2a\lim_{h\to0}(2a+h)=2a

This leads to the power rule. You can differentiate a polynomial one term at a time; constant terms have zero slope.

ddxxn=nxn−1,ddx(u+v)=u′+v′,ddxc=0\frac{d}{dx}x^n=nx^{n-1},\qquad\frac{d}{dx}(u+v)=u'+v',\qquad\frac{d}{dx}c=0

Check your understanding

Choose an answer to see why it works. You can change your choice.

  1. Question 1

    What is the derivative of this function?

    f(x)=x2f(x)=x^2

  2. Question 2

    What is the tangent slope at this input?

    f(x)=x2,x=3f(x)=x^2,\quad x=3

  3. Question 3

    Differentiate each term and add the results.

    g(x)=3x2+2x−5g(x)=3x^2+2x-5