Part III · Differential Calculus and Local Approximation

Chapter 7 — Derivatives from First Principles

Why This Matters

An average speed over ten seconds does not tell us the speed at one instant. In the same way, the ratio $\Delta V/\Delta t$ describes an average voltage change over an interval, while a circuit or control problem often needs the instantaneous rate of change. The derivative is the limiting version of that average rate.

7.1 Average rate of change and secant slope

For a function $$y=f(x)$$ , the average change in output between $$x$$ and $$x+h$$ is

\frac{f(x+h)-f(x)}{h},\qquad h\ne0.

On a graph, the two points $$(x,f(x))$$ and $$(x+h,f(x+h))$$ determine a secant line. Its slope is precisely the displayed quotient. In a physical model where $$x=t$$ is time, it is average rate of change over a time interval of length $$h$$ .

As $$h$$ becomes smaller, the second point moves towards the first. If the secant slopes approach one definite number, that number is the slope of the tangent line and the instantaneous rate of change.

Average rate is a line through two measured points. Moving from x to x+h changes both input and output, so the quotient is the secant slope. In a time model it is a rate over an interval, not yet an instantaneous rate.

\frac{f(x+h)-f(x)}h,\qquad h\ne0

Two points on y equals x squared define a secant slope before the offset h is allowed to shrink.

As the secant offset h shrinks toward zero, the secant slope through (1, 1) and (1+h, (1+h)^2) approaches the tangent slope 2.

7.2 Definition of derivative

The derivative of $$f$$ at $$x$$ is

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h},

provided the limit exists. It is a new function when it can be computed at many inputs. Other notations include $\frac{df}{dx}$ and, when $$y=f(x)$$ , $\frac{dy}{dx}$ .

Example — Derivative of a square from the definition

Let $$f(x)=x^2$$ . Start with the difference quotient:

\frac{f(x+h)-f(x)}h=\frac{(x+h)^2-x^2}{h}.

Expand the square before cancelling:

\frac{x^2+2xh+h^2-x^2}{h}=\frac{h(2x+h)}h=2x+h.

The cancellation is valid because the quotient is considered for $h\ne0$ . Now take the limit:

f'(x)=\lim_{h\to0}(2x+h)=2x.

At $$x=3$$ , the tangent slope is $$6$$ . This result is not guessed from the graph; the graph is an interpretation of the limiting calculation.

Expand before taking the limit. For a square, the non-zero offset permits cancellation only after expansion produces a factor of h. The simplified secant slope 2x+h then has a limit of 2x, which is the tangent slope.

\frac{(x+h)^2-x^2}{h}=2x+h\longrightarrow2x

Keep h non-zero while forming the difference quotient.
Expand and factor h before cancelling it.
Only then let h tend to zero to obtain the derivative.

As h tends to zero at x equals 1, the secant slope of x squared tends to the tangent slope 2.

Stepping h from 2 toward 0, the secant slope 2x+h at x=1 approaches the derivative 2; the two points merge into the tangent line.

7.3 When a derivative does not exist

Differentiability is stronger than continuity. A differentiable function is continuous, but a continuous function need not be differentiable.

For $$f(x)=|x|$$ , the left-hand slopes at $$0$$ are $$-1$$ and the right-hand slopes are $$1$$ . Because these do not agree, $$f'(0)$$ does not exist. A jump discontinuity also prevents differentiability. A graph may additionally have a vertical tangent, where the ordinary finite derivative fails to exist.

Continuity does not settle the slope question. At the corner of |x|, left slopes are negative one and right slopes are positive one, so no single derivative exists. Jumps and vertical tangents likewise fail to supply an ordinary finite tangent slope.

\lim_{h\to0^-}\frac{|h|}{h}=-1\ne1=\lim_{h\to0^+}\frac{|h|}{h}

7.4 Linearity rules

The derivative respects addition and constant scaling:

\frac d{dx}[cf(x)]=cf'(x), \qquad \frac d{dx}[f(x)+g(x)]=f'(x)+g'(x).

These follow from the corresponding limit laws. The derivative of a constant is $$0$$ , because its output does not change, and $\frac d{dx}x=1$ .

Local increments add and scale before their limit is taken. The difference quotient of a sum splits into two quotients, and a constant factor stays attached to its changing output. Constants differentiate to zero because their local increment is zero.

(cf+g)'=cf'+g',\qquad\frac d{dx}C=0

7.5 Product and quotient rules

For a product, both factors may change. Add and subtract the intermediate term $$f(x+h)g(x)$$ :

\begin{aligned} &f(x+h)g(x+h)-f(x)g(x)\\ ={}&f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)]. \end{aligned}

Divide by $$h$$ and take the limit. Continuity of differentiable functions lets $f(x+h)\to f(x)$ , yielding

\[(fg)'=f'g+fg'.\]

For $g(x)\ne0$ , write $f/g=f\,g^{-1}$ and differentiate, or derive directly:

\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}.

Example — Why the product rule needs two terms

For $$P(t)=v(t)i(t)$$ , electrical power changes because voltage can change, current can change, or both can change:

\frac{dP}{dt}=v\frac{di}{dt}+i\frac{dv}{dt}.

Keeping only one term would assume, without stating it, that the other factor is constant.

Two changing factors create two contributions. Add and subtract the intermediate product f(x+h)g(x) to split the total change into a change of g and a change of f. The quotient rule inherits the additional condition that the denominator remains non-zero.

(fg)'=f'g+fg',\qquad\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}\ (g\ne0)

Insert and subtract f of x plus h times g of x.
Group the two resulting local increments and divide by h.
Take the limit while retaining the quotient domain restriction where needed.

7.6 Chain rule

Suppose $$y=f(u)$$ and $$u=g(x)$$ . A small change in $$x$$ first changes $$u$$ , which then changes $$y$$ . The derivative is

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(g(x))g'(x).

This is the chain rule. It is a statement about composition, not a trick about parentheses.

Example — A nested power

Differentiate $$y=(3x^2-1)^5$$ . The outside function is $$u^5$$ and the inside is $$u=3x^2-1$$ . Therefore

\frac{dy}{dx}=5(3x^2-1)^4\cdot6x=30x(3x^2-1)^4.

A local change travels through each layer. A small change in x first changes the inner quantity and then changes the outer output. The chain rule multiplies those local rates in their physical order; it is not a rule for merely removing parentheses.

\frac{dy}{dx}=f'(g(x))g'(x)

Name the inner quantity u equals g of x.
Differentiate the outer rule with respect to u.
Multiply by du over dx to return the rate to x.

7.7 Power, polynomial, and rational functions

For a real exponent $$a$$ , the power rule

\frac d{dx}x^a=ax^{a-1}

is used on intervals where the function is real and differentiable. For arbitrary real $$a$$ , the standard safe domain is $$x>0$$ . Integer powers have their usual larger domains; selected rational powers may have additional real inputs.

Term-by-term differentiation gives, for example,

\frac d{dx}(4x^3-5x^2+7x-9)=12x^2-10x+7.

For rational functions, use the quotient rule or first simplify carefully while retaining all domain restrictions.

Rules inherit the original function's domain. The power rule applies where the real-valued power is defined and differentiable; arbitrary real exponents are safely handled on x>0. Simplifying a rational expression never grants a derivative at an input excluded by its original denominator.

\frac d{dx}x^a=ax^{a-1}\quad\text{on a real differentiable domain}

7.8 Higher derivatives

The second derivative is the derivative of the derivative:

f''(x)=\frac{d}{dx}f'(x).

In motion it measures acceleration; in a graph it records how the slope changes. For a voltage signal, $$v'(t)$$ and $$v''(t)$$ describe increasingly rapid aspects of variation. Higher derivatives later provide Taylor approximations and determine the form of second-order differential equations.

The next derivative tracks change in the current rate. If f′ is slope or instantaneous rate, then f″ measures how that slope itself changes. This connects acceleration, curvature, and the higher local information used later in Taylor approximation.

f''(x)=\frac d{dx}f'(x)

Common Mistakes

Cancelling $$h$$ before expanding a difference quotient or forgetting that it is nonzero before the limit is taken.
Using the product rule as $$(fg)'=f'g'$$ .
Forgetting the derivative of the inside function in a chain-rule calculation.
Applying $$x^a$$ formulas at points outside the real domain of the original function.
Inferring differentiability solely from a graph that has not been examined at the relevant scale.

Chapter Summary

The derivative begins as average change over an interval and becomes instantaneous change through a limit. Difference quotients justify the derivative definition. Linearity, product, quotient, chain, and power rules make derivatives practical, but their domain conditions remain part of every calculation.

7.9 Derivative rules as controlled local changes

The product rule has two terms because both factors can change. The chain rule has an inside derivative because the outer function receives a changing input. Before differentiating, identify the outermost operation and the independent variable; this prevents applying rules mechanically.

Example — quotient and domain together

For $$f(x)=(x^2+1)/(x-1)$$ , $x\ne1$ . The quotient rule gives

f'(x)=\frac{(2x)(x-1)-(x^2+1)}{(x-1)^2} =\frac{x^2-2x-1}{(x-1)^2}.

The derivative cannot be used to claim anything about $$x=1$$ , where the original function is not defined.

Example — higher derivatives describe changing slope

For $$s(t)=t^3-6t^2+9t$$ ,

s'(t)=3t^2-12t+9,\qquad s''(t)=6t-12.

The first derivative gives instantaneous rate; the second says whether that rate is increasing or decreasing. At $$t=2$$ , $$s''(2)=0$$ , so the slope changes its trend there. Later, this same information supports curvature and Taylor approximations.

[Figure: A curve with two nearby secant lines converging to a tangent; show a corner where left/right slopes differ and a smooth point with one tangent.]

Name the changing pieces before selecting a rule. A product has two independently changing factors, a composition has an inner scale, and a quotient retains an excluded denominator. Higher derivatives then give a second layer of rate information rather than a new unrelated calculation.

s(t)=t^3-6t^2+9t\Longrightarrow s'(t)=3t^2-12t+9,\ s''(t)=6t-12

Exercises

Use the limit definition to find the derivative of $$f(x)=3x-2$$ .
Explain why $$|x|$$ is continuous but not differentiable at $$0$$ .
Differentiate $$(x^2+1)(x^3-4)$$ .
Differentiate $\sqrt{1+4x^2}$ on its real domain.
If $$q(t)=C v(t)$$ for a capacitor with constant capacitance $$C$$ , derive a formula for $$dq/dt$$ .
Differentiate $$(2x-1)^4/(x+1)$$ and state its domain.
A signal is $$v(t)=t^3-6t^2+9t$$ . Find $$v'$$ and $$v''$$ , then locate points where its slope is zero.