3 The Derivative and Differentiation Rules

A progressive guide to defining derivatives, interpreting instantaneous change, and applying the main differentiation rules to explicit and implicit functions.

From Average Change to the

The central idea is instantaneous change. For a function ff, the average rate of change from x=ax=a to x=a+hx=a+h is

f(a+h)−f(a)h,h≠0.\frac{f(a+h)-f(a)}{h},\qquad h\ne 0.

This is the slope of a secant line through two points on the graph. Letting hh approach zero moves the second point toward the first and produces the slope of the :

f′(a)=lim⁡h→0f(a+h)−f(a)h.f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.

Replacing aa with the variable xx gives the function:

f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.

A limit-based example

For f(x)=x2f(x)=x^2, simplify the before taking the limit:

f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→02xh+h2h=lim⁡h→0(2x+h)=2x.\begin{aligned} f'(x)&=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\ &=\lim_{h\to 0}\frac{2xh+h^2}{h}\\ &=\lim_{h\to 0}(2x+h)\\ &=2x. \end{aligned}

Direct substitution into the unsimplified quotient would produce the indeterminate form 00\frac{0}{0}, so algebraic simplification comes first.

Takeaway: The is the limit of average rates of change as the input interval shrinks to zero.

Meaning and Notation

The has both geometric and applied meanings. At the point (a,f(a))(a,f(a)), f′(a)f'(a) is the slope of the . A positive indicates that the function is increasing locally, a negative indicates local decrease, and a zero gives a horizontal tangent.

If y=f(x)y=f(x), common notations are f′(x)f'(x), y′y', dydx\frac{dy}{dx}, and ddx[f(x)]\frac{d}{dx}[f(x)]. The notation f′(a)f'(a) means the value of the at one input, whereas f′(x)f'(x) denotes the whole function.

Derivatives also describe rates with units. If s(t)s(t) is position as a function of time, then

v(t)=s′(t)v(t)=s'(t)

is instantaneous velocity. The units of a are output units divided by input units.

Differentiability is a stronger condition than continuity: differentiability at a point implies continuity there, but continuity does not always imply differentiability. The function f(x)=∣x∣f(x)=|x| is continuous at x=0x=0, yet it has different one-sided slopes and is not differentiable at that point.

Takeaway: Interpret a as a slope, an instantaneous rate, and a quantity whose units reflect the ratio of output change to input change.

Core Differentiation Rules

The basic rules turn limit definitions into efficient calculations.

Core algebraic rules

For a constant CC,

ddx[C]=0.\frac{d}{dx}[C]=0.

For a constant multiple,

ddx[Cf(x)]=Cf′(x).\frac{d}{dx}[Cf(x)]=Cf'(x).

Sums and differences can be differentiated term by term:

ddx[f(x)+g(x)]=f′(x)+g′(x),\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x),
ddx[f(x)−g(x)]=f′(x)−g′(x).\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x).

The gives

ddx[xn]=nxn−1.\frac{d}{dx}[x^n]=nx^{n-1}.

For instance,

ddx[3x4−2x+7]=12x3−2.\frac{d}{dx}[3x^4-2x+7]=12x^3-2.

Products and quotients

Use the for a product:

ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x).\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x).

Use the quotient rule for a quotient with g(x)≠0g(x)\ne 0:

ddx[f(x)g(x)]=g(x)f′(x)−f(x)g′(x)[g(x)]2.\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}.

For example,

ddx[x2+1x−1]=(x−1)(2x)−(x2+1)(1)(x−1)2.\frac{d}{dx}\left[\frac{x^2+1}{x-1}\right]=\frac{(x-1)(2x)-(x^2+1)(1)}{(x-1)^2}.

Takeaway: First identify the outer algebraic structure—constant, sum, product, or quotient—then apply the matching rule carefully.

Common Formulas

Several formulas serve as a reference set:

  • ddx[ex]=ex\frac{d}{dx}[e^x]=e^x

  • ddx[ax]=axln⁡a\frac{d}{dx}[a^x]=a^x\ln a, where a>0a>0 and a≠1a\ne 1

  • ddx[ln⁡x]=1x\frac{d}{dx}[\ln x]=\frac{1}{x}

  • ddx[sin⁡x]=cos⁡x\frac{d}{dx}[\sin x]=\cos x

  • ddx[cos⁡x]=−sin⁡x\frac{d}{dx}[\cos x]=-\sin x

  • ddx[tan⁡x]=sec⁡2x\frac{d}{dx}[\tan x]=\sec^2 x

  • ddx[cot⁡x]=−[csc⁡x]2\frac{d}{dx}[\cot x]=-[\csc x]^2

  • ddx[sec⁡x]=sec⁡xtan⁡x\frac{d}{dx}[\sec x]=\sec x\tan x

  • ddx[csc⁡x]=−[csc⁡x][cot⁡x]\frac{d}{dx}[\csc x]=-[\csc x][\cot x]

For logarithms with another base, use

log⁡ax=ln⁡xln⁡a,\log_a x=\frac{\ln x}{\ln a},

which gives

ddx[log⁡ax]=1xln⁡a.\frac{d}{dx}[\log_a x]=\frac{1}{x\ln a}.

Two useful inverse-trigonometric formulas are

ddx[arcsin⁡x]=11−x2,ddx[arctan⁡x]=11+x2.\frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}.

Takeaway: Recognizing a function from a standard table reduces computation to selecting the correct formula and checking its domain.

Compositions and the

A composition has one function inside another, such as sin⁡(x2)\sin(x^2), (3x−1)5(3x-1)^5, or e4x+2e^{4x+2}. The handles these nested structures. If h(x)=f(g(x))h(x)=f(g(x)), then

h′(x)=f′(g(x))g′(x).h'(x)=f'(g(x))g'(x).

A reliable process is:

  1. Identify the outer function and the inner function.

  2. Differentiate the outer function while keeping the inner expression unchanged.

  3. Multiply by the of the inner function.

For example,

ddx(3x2−5)4=4(3x2−5)3ddx(3x2−5)=24x(3x2−5)3.\begin{aligned} \frac{d}{dx}(3x^2-5)^4 &=4(3x^2-5)^3\frac{d}{dx}(3x^2-5)\\ &=24x(3x^2-5)^3. \end{aligned}

For a trigonometric composition,

ddx[sin⁡(5x3)]=cos⁡(5x3)⋅15x2=15x2cos⁡(5x3).\frac{d}{dx}[\sin(5x^3)]=\cos(5x^3)\cdot 15x^2=15x^2\cos(5x^3).

In Leibniz notation, if y=f(u)y=f(u) and u=g(x)u=g(x), then

dydx=dydududx.\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.

Takeaway: Nested functions require both the of the outside layer and the of the inside layer.

Implicit Relationships and Slopes

Use when an equation relates xx and yy without isolating yy. Differentiate both sides with respect to xx, treat yy as a function of xx, collect terms containing dydx\frac{dy}{dx}, and solve.

Circle example

For

x2+y2=25,x^2+y^2=25,

differentiate both sides:

2x+2ydydx=0.2x+2y\frac{dy}{dx}=0.

Solving gives

dydx=−xy.\frac{dy}{dx}=-\frac{x}{y}.

At (3,4)(3,4), the slope is

dydx∣(3,4)=−34.\left.\frac{dy}{dx}\right|_{(3,4)}=-\frac{3}{4}.

The is therefore

y−4=−34(x−3).y-4=-\frac{3}{4}(x-3).

Product involving the dependent variable

For

xy+y2=6,xy+y^2=6,

differentiate using the and the :

xdydx+y+2ydydx=0.x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0.

Group the terms and solve:

(x+2y)dydx=−y,(x+2y)\frac{dy}{dx}=-y,
dydx=−yx+2y.\frac{dy}{dx}=-\frac{y}{x+2y}.

Takeaway: Every occurrence of a dependent variable such as yy contributes a factor of dydx\frac{dy}{dx} when differentiated with respect to xx.