04 The Derivative

A progressive guide to understanding the derivative as a rate of change, finding derivatives with core rules, and applying them to tangent lines, functions, and real-world quantities.

Meaning of the

The describes change at a single input value. For a function ff, the at aa is defined by

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

provided the limit exists. The expression inside the limit is the . It compares the change in output, f(a+h)−f(a)f(a+h)-f(a), with the small input change hh.

Over a complete interval from x=ax=a to x=bx=b, the is

f(b)−f(a)b−a.\frac{f(b)-f(a)}{b-a}.

This is the slope of a secant line. When bb moves closer to aa, the secant line approaches the , and its slope approaches the , which is the .

For example, if a particle's position is s(t)s(t), then

s(b)−s(a)b−a\frac{s(b)-s(a)}{b-a}

is its average velocity between two times, while s′(a)s'(a) is its velocity at the exact time aa.

Takeaway: The is both a limiting slope and an .

Notation and Motion

Several notations represent the same :

f′(x),y′,dydx,ddx[f(x)].f'(x),\qquad y',\qquad \frac{dy}{dx},\qquad \frac{d}{dx}[f(x)].

The notation f′(x)f'(x) describes a new function, while f′(a)f'(a) means that this function has been evaluated at the particular input aa.

For position and motion, if s(t)s(t) is position, then velocity is

v(t)=s′(t)=dsdt,v(t)=s'(t)=\frac{ds}{dt},

and acceleration is

a(t)=v′(t)=s′′(t)=d2sdt2.a(t)=v'(t)=s''(t)=\frac{d^2s}{dt^2}.

Thus, the second measures how velocity changes. Units also carry meaning: if position is measured in meters and time in seconds, then velocity has units of meters per second and acceleration has units of meters per second squared.

Takeaway: Notation identifies what is being differentiated, which variable is changing, and whether the result is a function or a value at a particular point.

Tangent Lines and

If a function is differentiable at x=ax=a, its at (a,f(a))(a,f(a)) has slope f′(a)f'(a). The point-slope equation is

y−f(a)=f′(a)(x−a).y-f(a)=f'(a)(x-a).

For f(x)=x2f(x)=x^2, the is f′(x)=2xf'(x)=2x. At x=3x=3, the point on the curve is (3,9)(3,9), and the slope is f′(3)=6f'(3)=6. Therefore,

y−9=6(x−3),y-9=6(x-3),

which simplifies to

y=6x−9.y=6x-9.

When xx is close to aa, the gives the local linear approximation

f(x)≈f(a)+f′(a)(x−a).f(x)\approx f(a)+f'(a)(x-a).

This approximation is useful because a complicated curve and its have nearly the same values near the point of tangency.

A function is differentiable at a point when its exists there. requires a well-defined finite tangent slope. A discontinuity, hole, sharp corner, cusp, or vertical tangent can cause the to fail to exist. implies , but does not imply . For instance, f(x)=∣x∣f(x)=|x| is continuous at 00, but its left-hand slope is −1-1 and its right-hand slope is 11, so it has no at 00.

Takeaway: A determines the tangent-line slope, and is a stronger condition than .

Core Differentiation Rules

The basic differentiation rules make symbolic calculation efficient.

  • Constant rule:

    ddx[c]=0.\frac{d}{dx}[c]=0.
  • :

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

    ddx[cf(x)]=cf′(x).\frac{d}{dx}[cf(x)]=cf'(x).
  • Sum and difference rules:

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

For a polynomial, differentiate each term separately. For example,

f(x)=4x5−3x2+7x−9f(x)=4x^5-3x^2+7x-9

has

f′(x)=20x4−6x+7.f'(x)=20x^4-6x+7.

The constant term disappears because its is zero.

For products, use the :

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).

For quotients, use the :

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

For nested expressions, use the :

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

For example,

ddx[(3x2+1)5]=5(3x2+1)4(6x)=30x(3x2+1)4.\frac{d}{dx}[(3x^2+1)^5]=5(3x^2+1)^4(6x)=30x(3x^2+1)^4.

The factor 6x6x comes from differentiating the inner function 3x2+13x^2+1.

Takeaway: First identify the structure of an expression—sum, product, quotient, or composition—then select the corresponding rule.

Exponential and Logarithmic Derivatives

The exponential and logarithmic rules follow distinctive patterns. For the natural exponential function,

ddx[ex]=ex.\frac{d}{dx}[e^x]=e^x.

With a composite exponent,

ddx[eg(x)]=eg(x)g′(x).\frac{d}{dx}[e^{g(x)}]=e^{g(x)}g'(x).

For a general exponential function with a>0a>0 and a≠1a\ne 1,

ddx[ax]=axln⁡(a),\frac{d}{dx}[a^x]=a^x\ln(a),

and therefore

ddx[ag(x)]=ag(x)ln⁡(a)g′(x).\frac{d}{dx}[a^{g(x)}]=a^{g(x)}\ln(a)g'(x).

The natural logarithm satisfies

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

for x>0x>0. Applying the gives

ddx[ln⁡(g(x))]=g′(x)g(x),\frac{d}{dx}[\ln(g(x))]=\frac{g'(x)}{g(x)},

where the logarithm's input must be positive. For example,

ddx[ln⁡(x2+1)]=2xx2+1.\frac{d}{dx}[\ln(x^2+1)]=\frac{2x}{x^2+1}.

For a logarithm with base aa,

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

Takeaway: Exponential derivatives preserve the exponential factor, while logarithmic derivatives produce a reciprocal factor and require attention to domain.

Trigonometric Derivatives

The standard trigonometric formulas use radians as the angle measure:

ddx[sin⁡x]=cos⁡x,ddx[cos⁡x]=−sin⁡x,\frac{d}{dx}[\sin x]=\cos x, \qquad \frac{d}{dx}[\cos x]=-\sin x,
ddx[tan⁡x]=sec⁡2x,ddx[cot⁡x]=−csc⁡2x,\frac{d}{dx}[\tan x]=\sec^2 x, \qquad \frac{d}{dx}[\cot x]=-\csc^2 x,
ddx[sec⁡x]=sec⁡xtan⁡x,ddx[csc⁡x]=−csc⁡xcot⁡x.\frac{d}{dx}[\sec x]=\sec x\tan x, \qquad \frac{d}{dx}[\csc x]=-\csc x\cot x.

If the trigonometric function contains a composite argument, multiply by the of that argument. For example,

ddx[cos⁡(4x3−1)]=−sin⁡(4x3−1)⋅12x2=−12x2sin⁡(4x3−1).\frac{d}{dx}[\cos(4x^3-1)] =-\sin(4x^3-1)\cdot 12x^2 =-12x^2\sin(4x^3-1).

The inner function is 4x3−14x^3-1, whose is 12x212x^2. The supplies this factor.

Takeaway: Memorize the six basic trigonometric derivatives, use radians, and apply the to nontrivial arguments.

Interpreting and Applying Derivatives

A can be interpreted directly from its sign and magnitude.

  • If f′(x)>0f'(x)>0, the function is increasing locally.

  • If f′(x)<0f'(x)<0, the function is decreasing locally.

  • If f′(x)=0f'(x)=0, the is horizontal. This may correspond to a local maximum, a local minimum, or neither.

  • A large value of ∣f′(x)∣|f'(x)| indicates a steep graph and a rapid local rate of change.

The units of a are the units of the output divided by the units of the input. If C(q)C(q) is the cost in dollars of producing qq items, then C′(q)C'(q) has units of dollars per item. It approximates the additional cost associated with producing one more item near the production level qq.

A reliable workflow is:

  1. Identify the function and the variable with respect to which you are differentiating.

  2. Determine the expression's structure.

  3. Apply the appropriate rule, including the for compositions.

  4. Simplify the .

  5. Evaluate at a specified input if a numerical rate or tangent slope is required.

  6. Check the domain and interpret the units and sign.

Takeaway: Differentiation connects algebraic formulas with geometric slopes, physical motion, and practical rates of change.