Calculus · Chapter 3 — Differentiation Rules

3.1

Derivatives of Polynomials & Exponential Functions

The simplest functions have the simplest derivatives. Starting from constants and integer powers, a short list of rules covers every polynomial, and a single special base gives the cleanest exponential derivative in all of calculus.

Constant & Power Functions

Derivative of a Constant

A constant function has a horizontal graph, so its slope is zero everywhere. The rate of change of something that never changes is nothing.

Constant Function

\[ \frac{d}{dx}(c)=0 \]

Rule

The Power Rule. For any real exponent \(n\), the derivative of \(x^n\) is found by pulling the exponent down as a coefficient and reducing it by one. It holds first for positive integers, then extends to negative and fractional exponents, and finally to all real numbers.

Power Rule (General)

\[ \frac{d}{dx}\left(x^{n}\right)=n\,x^{\,n-1} \]

FIG 3.1.1A constant has slope 0; the line \(y=x\) has slope 1. These are the base cases of the Power Rule. (Original redrawn figure.)

Building New Derivatives From Old

Constant Multiple, Sum & Difference Rules

Differentiation is linear: a constant factor passes straight through, and the derivative of a sum or difference is the sum or difference of the derivatives. Combined with the Power Rule, this differentiates any polynomial term-by-term.

Linearity of the Derivative

\[ \frac{d}{dx}\!\left[c\,f(x)\right]=c\,f'(x) \] \[ \frac{d}{dx}\!\left[f(x)\pm g(x)\right]=f'(x)\pm g'(x) \]

EXAMPLE 1 Differentiating a polynomial

Differentiate \( x^{8}+12x^{5}-4x^{4}+10x^{3}-6x+5 \).

Solution

Apply the Sum, Difference, and Constant Multiple Rules to split the expression into single terms.

Differentiate each power with the Power Rule; the constant 5 contributes 0.

Result: \( 8x^{7}+60x^{4}-16x^{3}+30x^{2}-6 \).

EXAMPLE 2 Horizontal tangents

Find where the curve \( y=x^{4}-6x^{2}+4 \) has a horizontal tangent.

Solution

Differentiate: \( y'=4x^{3}-12x=4x\,(x^{2}-3) \).

A horizontal tangent needs \( y'=0 \), so \( x=0 \) or \( x=\pm\sqrt{3} \).

The curve is flat at \((0,4)\), \((\sqrt3,-5)\), and \((-\sqrt3,-5)\).

The Natural Exponential Function

Why \(e\) Is Special

The derivative of \(a^{x}\) is proportional to \(a^{x}\) itself, with a proportionality constant that depends on the base. The number \(e\) is defined as the base for which that constant is exactly 1 — making \(e^{x}\) its own derivative.

Natural Exponential

\[ \frac{d}{dx}\left(e^{x}\right)=e^{x} \]

The fact that \(e^{x}\) reproduces itself under differentiation is exactly why it governs growth, decay, and oscillating systems. When you see \(e\) in a model, it usually means "rate of change proportional to current amount" — keep that intuition handy for 3.8.

3.2

The Product & Quotient Rules

The derivative of a product is not the product of the derivatives — a tempting guess that fails on simple examples. Two correct rules handle functions joined by multiplication and division.

The Product Rule

The derivative of a product of two functions equals the first function times the derivative of the second, plus the second function times the derivative of the first. Each factor takes a turn being differentiated while the other is held fixed.

Product Rule

\[ \frac{d}{dx}\!\left[f(x)\,g(x)\right]=f(x)\,g'(x)+g(x)\,f'(x) \]

FIG 3.2.1The product \(uv\) as an area: a small change in each side adds two strips (\(u\,\Delta v\) and \(v\,\Delta u\)) plus a vanishing corner. (Original redrawn figure.)

The Quotient Rule

The derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. Order matters because of the minus sign.

Quotient Rule

\[ \frac{d}{dx}\!\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)\,f'(x)-f(x)\,g'(x)}{\left[g(x)\right]^{2}} \]

EXAMPLE 1 Product Rule with an exponential

Find \(f'(x)\) for \( f(x)=x\,e^{x} \).

Solution

Take \(f=x\), \(g=e^{x}\): \( f'(x)=x\dfrac{d}{dx}(e^{x})+e^{x}\dfrac{d}{dx}(x) \).

Simplify: \( f'(x)=x e^{x}+e^{x}=(x+1)e^{x} \).

EXAMPLE 2 Quotient Rule on a rational function

Differentiate \( y=\dfrac{x^{2}+x-2}{x^{3}+6} \).

Solution

Numerator derivative \(2x+1\); denominator derivative \(3x^{2}\).

Apply the rule and expand the numerator over \((x^{3}+6)^{2}\).

Result: \( y'=\dfrac{-x^{4}-2x^{3}+6x^{2}+12x+6}{(x^{3}+6)^{2}} \).

Don't reach for the Quotient Rule reflexively. If a quotient simplifies first — e.g. \(\dfrac{3x^{2}+2\sqrt{x}}{x}=3x+2x^{-1/2}\) — rewrite, then use the Power Rule. Fewer steps, fewer sign errors.

3.3

Derivatives of Trigonometric Functions

Reading slopes off the sine curve already hints that its derivative is the cosine. A careful limit argument confirms it, and the Quotient Rule delivers the rest of the family. These hold only when angles are measured in radians.

The Fundamental Trig Limit

As an angle shrinks toward zero, the ratio of its sine to its size approaches 1. This single limit is what powers the derivative of the sine function.

Key Limit

\[ \lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1 \qquad \lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}=0 \]

FIG 3.3.1The slope of \(\sin x\) at each point traces out \(\cos x\). (Original redrawn figure.)

Formulas

Derivatives of the six trig functions. The cofunctions (cosine, cosecant, cotangent) all carry a minus sign — a handy memory aid.

Trigonometric Derivatives

\(\dfrac{d}{dx}(\sin x)=\cos x\)

\(\dfrac{d}{dx}(\csc x)=-\csc x\cot x\)

\(\dfrac{d}{dx}(\cos x)=-\sin x\)

\(\dfrac{d}{dx}(\sec x)=\sec x\tan x\)

\(\dfrac{d}{dx}(\tan x)=\sec^{2}x\)

\(\dfrac{d}{dx}(\cot x)=-\csc^{2}x\)

EXAMPLE 1 Simple harmonic motion

An object on a spring has position \( s=4\cos t \). Find its velocity and acceleration.

Solution

Velocity: \( v=\dfrac{ds}{dt}=-4\sin t \).

Acceleration: \( a=\dfrac{dv}{dt}=-4\cos t \).

Acceleration is always \(-s\): the hallmark of simple harmonic motion. Speed peaks at equilibrium, where \(\cos t=0\).

Radians are not optional here. If you ever differentiate trig functions in degrees, an extra factor of \(\pi/180\) appears and every formula above breaks. In engineering work, keep angular quantities in radians end-to-end.

3.4

The Chain Rule

Composite functions — a function inside another — are everywhere. The Chain Rule is the single most important differentiation rule for applied work: it tells you how rates multiply through a chain of dependencies.

The Chain Rule

To differentiate a composition, multiply the derivative of the outer function (evaluated at the inner function) by the derivative of the inner function. Intuitively, if \(y\) changes \(k\) times as fast as \(u\), and \(u\) changes \(m\) times as fast as \(x\), then \(y\) changes \(km\) times as fast as \(x\).

Chain Rule

\[ \frac{d}{dx}f\!\left(g(x)\right)=f'\!\left(g(x)\right)\cdot g'(x) \] \[ \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \]

Power Rule + Chain Rule

\[ \frac{d}{dx}\!\left[g(x)\right]^{n}=n\left[g(x)\right]^{\,n-1}g'(x) \]

General Exponential Base

\[ \frac{d}{dx}\left(a^{x}\right)=a^{x}\ln a \qquad \frac{d}{dx}\left(e^{u}\right)=e^{u}\frac{du}{dx} \]

EXAMPLE 1 Outer–inner identification

Differentiate \( F(x)=\sqrt{x^{2}+1} \).

Solution

Outer: square root; inner: \( x^{2}+1 \).

\( F'(x)=\dfrac{1}{2\sqrt{x^{2}+1}}\cdot 2x=\dfrac{x}{\sqrt{x^{2}+1}} \).

EXAMPLE 2 Chained twice

Differentiate \( y=e^{\sin x} \) and \( y=\sin(x^{2}) \).

Solution

\( \dfrac{d}{dx}e^{\sin x}=e^{\sin x}\cos x \) — exponential outer, sine inner.

\( \dfrac{d}{dx}\sin(x^{2})=\cos(x^{2})\cdot 2x \) — sine outer, square inner.

Work strictly outside-in and don't skip the inner derivative — forgetting the \(g'(x)\) factor is the single most common differentiation mistake. For deep nests (a function of a function of a function), peel one layer at a time.

3.5

Implicit Differentiation

Some curves can't be solved cleanly for \(y\) in terms of \(x\). Implicit differentiation lets you find \(dy/dx\) directly from the equation, treating \(y\) as an unknown function of \(x\) and applying the Chain Rule wherever \(y\) appears.

The Method

Differentiate both sides of the equation with respect to \(x\), remembering that every \(y\) carries a hidden \(dy/dx\) factor by the Chain Rule. Then collect the \(dy/dx\) terms and solve for it algebraically.

EXAMPLE 1 Tangent to a circle

Find \(dy/dx\) for \( x^{2}+y^{2}=25 \).

Solution

Differentiate: \( 2x+2y\dfrac{dy}{dx}=0 \).

Solve: \( \dfrac{dy}{dx}=-\dfrac{x}{y} \). At \((3,4)\) the slope is \(-\tfrac34\).

Derivatives of Inverse Trig Functions

Inverses via Implicit Differentiation

Writing \(y=\sin^{-1}x\) as \(\sin y = x\) and differentiating implicitly produces clean algebraic derivatives for the inverse trigonometric functions — no transcendental functions remain.

Inverse Trigonometric Derivatives

\(\dfrac{d}{dx}(\sin^{-1}x)=\dfrac{1}{\sqrt{1-x^{2}}}\)

\(\dfrac{d}{dx}(\csc^{-1}x)=-\dfrac{1}{x\sqrt{x^{2}-1}}\)

\(\dfrac{d}{dx}(\cos^{-1}x)=-\dfrac{1}{\sqrt{1-x^{2}}}\)

\(\dfrac{d}{dx}(\sec^{-1}x)=\dfrac{1}{x\sqrt{x^{2}-1}}\)

\(\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{1}{1+x^{2}}\)

\(\dfrac{d}{dx}(\cot^{-1}x)=-\dfrac{1}{1+x^{2}}\)

Implicit differentiation is the engineer's shortcut for constraint equations — gear meshes, four-bar linkages, pressure-volume relations. You rarely solve the constraint explicitly; you differentiate it as-is and solve for the rate you want.

3.6

Derivatives of Logarithmic Functions

Logarithms are the inverses of exponentials, so implicit differentiation hands us their derivatives. The natural log gives the simplest formula, and "logarithmic differentiation" tames otherwise brutal products, quotients, and powers.

Logarithmic Derivatives

\[ \frac{d}{dx}(\ln x)=\frac{1}{x} \qquad \frac{d}{dx}(\log_a x)=\frac{1}{x\ln a} \] \[ \frac{d}{dx}\ln|x|=\frac{1}{x} \qquad \frac{d}{dx}\big[\ln g(x)\big]=\frac{g'(x)}{g(x)} \]

Logarithmic Differentiation

For complicated products, quotients, and powers, take the natural log of both sides first. The Laws of Logarithms turn multiplication into addition and exponents into coefficients, so differentiating becomes far simpler — then multiply back by \(y\) to recover \(dy/dx\).

EXAMPLE 1 Taming a messy quotient

Differentiate \( y=\dfrac{x^{3/4}\sqrt{x^{2}+1}}{(3x+2)^{5}} \).

Solution

Take logs: \( \ln y=\tfrac34\ln x+\tfrac12\ln(x^{2}+1)-5\ln(3x+2) \).

Differentiate implicitly: \( \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{3}{4x}+\dfrac{x}{x^{2}+1}-\dfrac{15}{3x+2} \).

Multiply by \(y\) to finish. Far cleaner than nested Product and Quotient Rules.

The Number \(e\) as a Limit

\[ e=\lim_{x\to 0}(1+x)^{1/x}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n} \]

The general Power Rule for all real exponents is finally proved here, using logarithmic differentiation. Also note the four exponent–base cases: constant^constant, function^constant (Power Rule), constant^function (exponential rule), and function^function (needs logs).

3.7

Rates of Change in the Sciences

A derivative is an instantaneous rate of change, and that one idea wears many costumes across the sciences — velocity, density, current, reaction rate, growth rate, and marginal cost are all the same mathematical object.

Velocity & Acceleration

For a particle with position \(s=f(t)\), velocity is the first derivative \(v=ds/dt\) and acceleration is the second derivative \(a=dv/dt\). A particle speeds up when velocity and acceleration share a sign, and slows down when they differ.

EXAMPLE 1 Analyzing motion

A particle moves with \( s=t^{3}-6t^{2}+9t \) (meters, seconds). Find when it is at rest.

Solution

Velocity: \( v=3t^{2}-12t+9=3(t-1)(t-3) \).

At rest when \(v=0\): \( t=1\,\text{s} \) and \( t=3\,\text{s} \).

Acceleration \( a=6t-12 \); the particle reverses direction between these times.

Other Interpretations

The same derivative becomes linear density \(\rho=dm/dx\) for a non-uniform rod, current \(I=dQ/dt\) for flowing charge, reaction rate \(d[\text{C}]/dt\) in chemistry, population growth rate \(dn/dt\) in biology, and marginal cost \(dC/dx\) in economics. One concept, developed once, applies everywhere.

Marginal Cost

\[ \text{marginal cost}=\frac{dC}{dx}\approx C(n+1)-C(n) \]

3.8

Exponential Growth & Decay

When a quantity changes at a rate proportional to its own size, the result is exponential. This single differential equation models population growth, radioactive decay, cooling, and compound interest.

Theorem

Law of Natural Growth / Decay. The only solutions of the differential equation \( dy/dt=ky \) are exponential functions \( y(t)=y(0)e^{kt} \). A positive \(k\) gives growth; a negative \(k\) gives decay. The constant \(k\) is the relative rate of change.

Governing Equation & Solution

\[ \frac{dy}{dt}=ky \quad\Longrightarrow\quad y(t)=y(0)\,e^{kt} \]

EXAMPLE 1 Radioactive half-life

Radium-226 has a half-life of 1590 years. Model a 100 mg sample.

Solution

\( m(t)=100\,e^{kt} \); use \( m(1590)=50 \) to find \( k=-\dfrac{\ln 2}{1590} \).

So \( m(t)=100\,e^{-(\ln 2)t/1590}=100\cdot 2^{-t/1590} \).

Newton's Law of Cooling

An object's temperature changes at a rate proportional to the difference between it and its surroundings. The substitution \( y=T-T_{s} \) reduces this to the natural-growth equation, so the temperature difference decays exponentially toward zero.

Continuously Compounded Interest

\[ A(t)=A_{0}\,e^{rt} \qquad \frac{dA}{dt}=rA(t) \]

Cooling, charging capacitors, settling transients, and first-order control systems all share the form \(dy/dt=k(y-y_\infty)\). Recognize it once and you can write the solution by inspection — no re-derivation needed.

3.9

Related Rates

When two changing quantities are linked by an equation, their rates of change are linked too. Related-rates problems compute a hard-to-measure rate from an easy-to-measure one — a workhorse technique for mechanism and flow analysis.

The Strategy

Find an equation relating the quantities, then differentiate both sides with respect to time using the Chain Rule. Substitute the known instantaneous values only after differentiating — never before, or you'll lose the very dependence you need.

EXAMPLE 1 Inflating balloon

Air enters a sphere at \( 100\ \text{cm}^3/\text{s} \). How fast is the radius growing when the diameter is 50 cm?

Solution

Relation: \( V=\tfrac43\pi r^{3} \). Differentiate: \( \dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt} \).

Solve for the unknown rate: \( \dfrac{dr}{dt}=\dfrac{1}{4\pi r^{2}}\dfrac{dV}{dt} \).

At \( r=25 \): \( \dfrac{dr}{dt}=\dfrac{100}{4\pi(25)^{2}}=\dfrac{1}{25\pi}\approx 0.0127\ \text{cm/s} \).

EXAMPLE 2 Sliding ladder

A 10 ft ladder slides; its base moves out at 1 ft/s. How fast does the top fall when the base is 6 ft out?

Solution

Pythagoras: \( x^{2}+y^{2}=100 \). Differentiate: \( 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0 \).

At \( x=6,\ y=8 \): \( \dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}=-\dfrac34 \) ft/s — the top slides down at \( \tfrac34 \) ft/s.

FIG 3.9.1Ladder against a wall: \(x\) and \(y\) both vary with time, linked by the Pythagorean Theorem. (Original redrawn figure.)

The golden rule: differentiate first, substitute numbers last. If you plug in fixed values too early, the changing quantity becomes a constant and its derivative wrongly drops to zero.

3.10

Linear Approximations & Differentials

Near a point, a smooth curve looks like its tangent line. That observation turns the tangent into a cheap, accurate stand-in for the function — the basis of countless engineering approximations.

Linearization

The tangent line at \(x=a\) approximates the function for \(x\) near \(a\). This linearization \(L(x)\) is the best first-degree approximation, accurate over a small interval and trivial to compute.

Linear Approximation

\[ f(x)\approx f(a)+f'(a)\,(x-a)=L(x) \]

EXAMPLE 1 Approximating a root

Linearize \( f(x)=\sqrt{x+3} \) at \( a=1 \) and estimate \(\sqrt{4.05}\).

Solution

\( f(1)=2,\ f'(1)=\tfrac14 \), so \( L(x)=\tfrac74+\tfrac{x}{4} \).

\( \sqrt{4.05}\approx \tfrac74+\tfrac{1.05}{4}=2.0125 \) — a slight overestimate, since the tangent lies above the curve.

Differentials

The differential \(dy=f'(x)\,dx\) measures the change along the tangent line, while \(\Delta y\) is the true change along the curve. For small \(dx\) they nearly agree, which makes \(dy\) a convenient estimate of error in a measured quantity.

Differential & Useful Approximations

\[ dy=f'(x)\,dx \qquad \sin\theta\approx\theta,\quad \cos\theta\approx 1 \;\; (\theta\text{ small}) \]

The small-angle approximations \(\sin\theta\approx\theta\) and \(\cos\theta\approx 1\) underpin pendulum analysis and paraxial optics. Differentials also give the standard error-propagation shortcut: relative error in a volume is about three times the relative error in the radius.

3.11

Hyperbolic Functions

Certain combinations of \(e^{x}\) and \(e^{-x}\) appear so often they earn their own names. The hyperbolic functions mirror the trig functions and describe hanging cables, ocean waves, and damped phenomena.

Definitions

Hyperbolic sine and cosine are the odd and even parts of the exponential function. They satisfy the identity \( \cosh^{2}x-\sinh^{2}x=1 \), the hyperbolic analog of the Pythagorean identity — which is why the point \((\cosh t,\sinh t)\) traces a hyperbola.

Definitions

\[ \sinh x=\frac{e^{x}-e^{-x}}{2}\qquad \cosh x=\frac{e^{x}+e^{-x}}{2}\qquad \tanh x=\frac{\sinh x}{\cosh x} \]

Hyperbolic Derivatives

\(\dfrac{d}{dx}(\sinh x)=\cosh x\)

\(\dfrac{d}{dx}(\operatorname{csch} x)=-\operatorname{csch} x\coth x\)

\(\dfrac{d}{dx}(\cosh x)=\sinh x\)

\(\dfrac{d}{dx}(\operatorname{sech} x)=-\operatorname{sech} x\tanh x\)

\(\dfrac{d}{dx}(\tanh x)=\operatorname{sech}^{2}x\)

\(\dfrac{d}{dx}(\coth x)=-\operatorname{csch}^{2}x\)

FIG 3.11.1A hanging cable settles into a catenary, the graph of a hyperbolic cosine. (Original redrawn figure.)

Inverse Hyperbolic (Logarithmic Forms)

\[ \sinh^{-1}x=\ln\!\left(x+\sqrt{x^{2}+1}\right)\qquad \cosh^{-1}x=\ln\!\left(x+\sqrt{x^{2}-1}\right) \]

The catenary \(y=c+a\cosh(x/a)\) is the real shape of suspended cables and chains — not a parabola, though it looks similar. The Gateway Arch is an inverted, weighted catenary. Useful whenever something flexible hangs under its own weight.