The Derivative

Part 2, Chapter 2: Single-Variable Real Analysis

Learning objectives

Define the derivative as $f'(a)=\lim_{h\to 0}(f(a+h)-f(a))/h$
Prove that differentiability implies continuity, but not conversely
State and apply the Mean Value Theorem and its sign-of-derivative corollaries
Use Taylor's theorem to approximate functions by polynomials

The derivative is the prototype of all linearisation. Real systems are rarely linear, but they are locally linear, that is precisely the geometric content of $f'(a)$ . Whenever you read about Newton's method, gradient descent, Kalman filters, control theory, or first-order asymptotic analysis, you are watching the derivative do its real job: providing the best linear approximation to a non-linear function at a point. The Mean Value Theorem (MVT) then turns local approximation into global conclusions, like "if the derivative is positive everywhere, the function is increasing".

The derivative as a limit

The derivative of $f$ at $a$ is $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ hto0fracf(a+h)−f(a)h, provided this limit exists. Geometrically it is the slope of the tangent line; physically it is an instantaneous rate of change. The fraction $(f(a+h)-f(a))/h$ is the difference quotient, the slope of the secant from $(a,f(a))$ to $(a+h,f(a+h))$ . Differentiation is the operation of taking the limit of secant slopes.

Differentiability implies continuity (not conversely)

If $f'(a)$ exists, then $f(x)-f(a)=(x-a)\cdot\frac{f(x)-f(a)}{x-a}$ . As $x\to a$ , the second factor approaches $f'(a)$ (finite) and the first factor approaches 0, so $f(x)\to f(a)$ . The converse fails dramatically: $f(x)=|x|$ is continuous everywhere but has $f'(0^-)=-1$ and $f'(0^+)=+1$ , so $f'(0)$ does not exist. There exist functions (Weierstrass's "monster") that are continuous everywhere but differentiable nowhere, analysis is full of such surprises.

The Mean Value Theorem

The Mean Value Theorem (MVT) says: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then there exists $c\in(a,b)$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$ . Geometrically: somewhere in the interval, the tangent line has the same slope as the chord connecting the endpoints. From MVT we instantly conclude: if $f'\equiv 0$ , then $f$ is constant; if $f'>0$ , then $f$ is strictly increasing; and the more subtle Cauchy MVT underpins L'Hôpital's rule.

Use the grapher to plot $f(x)=x^2$ on $[1,3]$ . The chord slope is $(9-1)/(3-1)=4$ , and the tangent has slope $f'(x)=2x$ . MVT predicts $c$ with $2c=4$ , i.e. $c=2$ . Visually you can see the tangent at $x=2$ is parallel to the chord. Try $x^3$ on the same interval to see a different $c$ , and try $|x|$ on $[-1,1]$ to see what happens when differentiability fails at an interior point (MVT does not apply).

Taylor's theorem

Taylor's theorem extends MVT: if $f$ is $n+1$ times differentiable, then $f(a+h)=f(a)+f'(a)h+\frac{f''(a)}{2!}h^2+\cdots+\frac{f^{(n)}(a)}{n!}h^n+R_n(h)$ n(h), where the remainder $R_n(h)=\frac{f^{(n+1)}(\xi)}{(n+1)!}h^{n+1}$ n(h)=fracf(n+1)(xi)(n+1)!hn+1 for some $\xi$ between $a$ and $a+h$ . This is why polynomials of high degree can approximate smooth functions with arbitrary accuracy, the whole basis of numerical analysis.

Where this shows up

Newton's laws & physics: velocity is the derivative of position, acceleration the derivative of velocity, force is mass times acceleration. Maxwell's equations, Schrödinger's equation, and the Einstein field equations are all differential equations, equations involving derivatives.
Gradient descent & machine learning: every neural-network parameter update is $\theta \leftarrow \theta - \eta \nabla L(\theta)$ . Backpropagation is the chain rule executed at industrial scale. Without the Mean Value Theorem and Taylor expansions, you cannot prove that gradient descent converges to a local minimum.
Control theory & robotics: PID controllers, Kalman filters, and model-predictive control all linearise non-linear dynamics around an operating point using $f'(a)$ (or its multivariable analogue, the Jacobian). The "first-order Taylor expansion" is the engineer's daily bread.

Pause and think: A function $f$ is differentiable at $0$ with $f'(0)=2$ . What is $\lim_{x\to 0}f(x)$ xto0f(x)? Justify in one line using the implication "differentiable implies continuous".

Try it

Predict first: use the limit definition to compute $f'(x)$ for $f(x)=1/x$ at $x=2$ . Hint: expand $1/(2+h)-1/2$ over a common denominator.
Apply MVT to $f(x)=\sin x$ on $[0,\pi/2]$ . Which $c$ does the theorem guarantee? Solve $\cos c = 2/\pi$ numerically.
Show that $f(x)=x|x|$ is differentiable at $0$ and find $f'(0)$ . (Use the limit definition; the answer is 0, the function is C^1 but not C^2 at 0.)
True or false: if $f'(x)>0$ for all $x$ in $(a,b)$ , then $f$ is strictly increasing on $(a,b)$ . Prove using MVT.

A trap to watch for

The implication "differentiable implies continuous" is one-way only. Many beginners assume continuous implies differentiable, then are baffled when they meet $|x|$ , the Cantor staircase, or the Weierstrass nowhere-differentiable function. The right mental model: continuity controls vertical jumps; differentiability additionally controls corners and cusps.

What you now know

You can compute derivatives from the limit definition, prove the differentiable-implies-continuous implication, apply MVT and its corollaries, and understand Taylor expansions as the canonical local approximation. Section §2.4 (Integration) is the dual: instead of measuring rates of change, we measure accumulated change, and section §2.5 (FTC) shows the two operations are mutual inverses.

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References

Garrity, T. (2002). All the Mathematics You Missed. Cambridge University Press, ch. 2.
Spivak, M. (2008). Calculus (4th ed.). Publish or Perish, ch. 10-11.
Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill, ch. 5.
Abbott, S. (2015). Understanding Analysis (2nd ed.). Springer, ch. 5.
Apostol, T. M. (1974). Mathematical Analysis (2nd ed.). Addison-Wesley, ch. 5.