Derivatives and Differentiation
Connect instantaneous change to a limit, choose differentiation rules, and interpret tangent slopes with units and domains.
From nearby values to a rate
Limits describe where outputs are heading. A derivative applies that idea to a rate: shrink the interval over which you measure change and ask whether the average rates approach a finite value. This produces a local description of motion, growth, cost or any other changing quantity.
Learning goals
Connect secants to tangents; differentiate from the definition; choose and combine differentiation rules; state valid domains; and interpret slopes and tangent equations.
Average and instantaneous change
Average rate of change
For distinct inputs a and b, the average rate is \((f(b)-f(a))/(b-a)\). Its units are output units per input unit, and it is the slope of the secant through the two graph points.
Profit over a production interval
A workshop models profit by \(P(x)=45x-0.025x^2-5000\) dollars. Compare production at 800 and 850 items.
Show worked solution
P(800)=15,000 and P(850)=15,187.50 dollars. The total increase is $187.50; dividing by 50 items gives an average rate of $3.75 per item. Total change and rate of change answer different questions.
Do not confuse units
A reservoir rises from 120 to 165 cubic meters in 3 hours. Find its average rate of change.
Hint
Divide the volume change by elapsed time.
Show worked solution
The rate is (165−120)/3=15 cubic meters per hour. It does not assert that the rate was constant during the three hours.
Velocity and speed
For differentiable position s(t) along a line, velocity is s′(t), a signed rate. Speed is |s′(t)|. Acceleration, when it exists, is s′′(t).
Reversing direction
A trolley has position s(t)=t²−4t meters for t≥0. Describe its motion at t=1,2,3 seconds.
Show worked solution
Velocity is 2t−4, so the velocities are −2,0,2 m/s and the speeds are 2,0,2 m/s. The trolley moves in the negative direction, pauses instantaneously, then moves in the positive direction. Acceleration remains 2 m/s². Zero velocity at one instant does not mean it is stationary over a time interval.
The derivative as a limit
Derivative at a point
\(f'(a)=\lim_{h\to0}[f(a+h)-f(a)]/h\), provided the finite limit exists. The difference quotient is formed only for \(h\ne0\); its limit defines the instantaneous rate.
The two nearby points determine a secant. As the second point approaches the first, the secant slopes may settle to a tangent slope. We take a limit; we never substitute zero into an uncanceled difference quotient.
Differentiate from first principles
Find the derivative of \(f(x)=4x-x^2\).
Show worked solution
Thus f′(x)=4−2x. At x=1 the instantaneous rate is 2.
A square-root difference quotient
Find \(f'(x)\) for \(f(x)=\sqrt{x+2}\), using the definition for x>−2.
Hint
Multiply the numerator and denominator by the conjugate.
Show worked solution
The quotient becomes \(1/(\sqrt{x+h+2}+\sqrt{x+2})\). Taking h→0 gives \(f'(x)=1/(2\sqrt{x+2})\). The finite derivative formula excludes x=−2.
From secant to tangent
When a derivative does not exist
Differentiability
A function is differentiable at a point when its derivative exists as a finite number. It is differentiable on an open interval when this holds at every point of that interval.
A derivative demands more than a connected graph: the left and right rates must agree and remain finite. Discontinuities, corners and vertical tangents can prevent this.
Continuous but not differentiable
Test \(f(x)=|x|\) at zero.
Show worked solution
The graph is continuous at 0. The left difference quotients equal −1 and the right ones equal 1. Their limits disagree, so f′(0) does not exist.
Keep in mind
Differentiability implies continuity, but continuity alone does not imply differentiability. A vertical tangent is not a finite derivative.
A flat tangent is allowed
Does \(f(x)=x^3\) have a derivative at zero?
Hint
Apply the limit to h³/h.
Show worked solution
The difference quotient is h², which tends to 0. A zero slope is a valid derivative; it is not a failure of differentiability.
Differentiable does not mean a continuous derivative
Let \(f(x)=x^2\sin(1/x)\) for x≠0 and f(0)=0. Does a derivative exist at zero?
Show worked solution
The difference quotient is h sin(1/h), whose absolute value is bounded by |h|, so f′(0)=0. For x≠0, f′(x)=2x sin(1/x)−cos(1/x). This derivative keeps oscillating near zero and is not continuous there. Differentiability of f and continuity of f′ are separate properties.
Power, constant and sum rules
Basic rules
\((C)'=0\), \((x^n)'=nx^{n-1}\), \((kf)'=kf'\), and \((f+g)'=f'+g'\). Apply these where the functions are defined and differentiable; for a general real exponent, x>0 is a sufficient domain.
Rewrite roots and reciprocals as powers before differentiating. The exponent changes by subtraction of one, not by dividing the exponent.
Rewrite before applying a rule
Differentiate \(y=5/(3x^2)-4\sqrt{x}\) for x>0.
Show worked solution
The original real domain requires x>0.
A derivative is a function
For \(f(x)=x^4-32x^2+10\), find f′(x) and f′(1).
Hint
Differentiate term by term, then evaluate.
Show worked solution
f′(x)=4x³−64x and f′(1)=−60. The first is a rule; the second is one slope.
Products and quotients
Product and quotient rules
\((uv)'=u'v+uv'\); when \(v\ne0\), \((u/v)'=(u'v-uv')/v^2\). Both component functions must be differentiable at the input.
A product changes because either factor can change. Differentiating only the factors and multiplying loses both contributions.
Use the quotient structure
Find the derivative of \(f(x)=(x^2+3)/(x+1)\).
Show worked solution
This holds for x≠−1. The squared denominator does not restore the excluded input.
Find the mistake
A student claims \((x^2\cdot x^3)'=2x\cdot3x^2=6x^3\). Correct the reasoning.
Hint
Expand the product first as an independent check.
Show worked solution
The product is x⁵, so the derivative is 5x⁴. The product rule also gives 2x·x³+x²·3x²=5x⁴. The proposed product of derivatives is not the product rule.
The chain rule tracks inner change
Chain rule
If u is differentiable at x and f is differentiable at u(x), then \(\frac{d}{dx}f(u(x))=f'(u(x))u'(x)\).
Differentiate the outer operation while holding its input as a single expression, then multiply by the rate of that input. Nested operations may require several factors.
A nested power
Differentiate \(y=(x^2+3x)^5\).
Show worked solution
The outer fifth power contributes the first factor; the inner quadratic contributes 2x+3.
Product and chain together
Differentiate \(y=-2x^5(2x-3x^2)^{1/3}\) wherever the factors are differentiable.
Hint
Keep the two product-rule terms and differentiate the inner quadratic.
Show worked solution
This rule-based expression applies when the inner expression is nonzero. Any excluded point requires a separate derivative check.
Do not lose the inner rate
Exponential and logarithmic rates
Exponential and logarithmic rules
\((e^u)'=e^u u'\); for a>0, \((a^u)'=a^u\ln(a)u'\). For u>0, \((\ln u)'=u'/u\). For a>0 and a≠1, \((\log_a u)'=u'/(u\ln a)\).
The logarithm base matters. Natural logarithms remove the extra base factor; other bases retain it. Always determine the real logarithm domain before simplifying.
Exponential composition
Differentiate \(f(x)=e^{3x^2+4}\).
Show worked solution
The exponential remains, multiplied by the derivative of its exponent.
Logarithm with a domain
Differentiate \(g(x)=\ln(x^2-1)\).
Show worked solution
The original logarithm requires x<−1 or x>1. A formula that can be evaluated elsewhere does not extend the original derivative domain.
A decay rate
For \(V(t)=2.5e^{-0.1t}\), find V′(5).
Hint
Include the derivative of the exponent.
Show worked solution
The negative rate means the modeled quantity is decreasing at t=5.
Logarithmic differentiation for a variable power
Differentiate \(y=x^x\) for x>0.
Show worked solution
Both the base and exponent vary, so neither the ordinary fixed-power rule nor the constant-base exponential rule applies alone. Take logarithms: ln y=x ln x. Differentiation gives y′/y=ln x+1. Thus y′=xˣ(ln x+1). The positive domain permits the logarithm.
A nested logarithm
Differentiate \(\ln(\ln x)\) on its real domain.
Hint
The outer logarithm requires ln x>0.
Show worked solution
Trigonometric derivatives in radians
Sine and cosine
For radian input, \((\sin u)'=\cos(u)u'\) and \((\cos u)'=-\sin(u)u'\). The radian convention is essential to these formulas.
Three nested operations
Differentiate \(f(x)=\cos(x\ln(\sin x))\).
Show worked solution
The outer cosine uses the chain rule; the inner product uses the product rule; the logarithm of sine uses another chain rule. The real domain requires sin x>0.
Keep in mind
Not every problem needs full algebraic expansion. A correctly factored derivative often makes the chain of reasoning easier to inspect.
Remaining trigonometric rules
\((\tan x)'=\sec^2x\), \((\cot x)'=-\csc^2x\), \((\sec x)'=\sec x\tan x\), and \((\csc x)'=-\csc x\cot x\), on their respective domains, with radian input. For a composite input u(x), multiply by u′(x).
These follow from sine and cosine derivatives using quotient or reciprocal rules. For example, differentiating sin x/cos x gives (cos²x+sin²x)/cos²x=sec²x wherever cos x≠0.
Tangent lines and horizontal tangents
Tangent line
If f is differentiable at a, its tangent line is \(y=f(a)+f'(a)(x-a)\). A horizontal tangent has \(f'(a)=0\).
Slope and point are both needed
Find the tangent to \(f(x)=(x^2+3)/(x+1)\) at x=2.
Show worked solution
The point is (2,7/3). The quotient-rule derivative gives f′(2)=5/9. Thus y−7/3=(5/9)(x−2).
Locate horizontal tangents
For the same function, find all horizontal tangent points.
Hint
Solve the numerator of f′=0 while retaining the domain.
Show worked solution
x²+2x−3=(x+3)(x−1)=0, so x=−3 or 1. Neither is excluded. The points are (−3,−6) and (1,2).
Inverse-function derivative
If f has a differentiable local inverse g near f(c) and f′(c)≠0, then g′(f(c))=1/f′(c).
Differentiate f(g(y))=y: the chain rule gives f′(g(y))g′(y)=1. In a graph reflection across y=x, horizontal and vertical changes swap, explaining the reciprocal slope. A zero original slope requires separate analysis.
Recover an inverse rate
A one-to-one function satisfies f(2)=7 and f′(2)=4. Find the derivative of its differentiable inverse at 7.
Show worked solution
The inverse takes 7 back to 2, so its derivative is 1/f′(2)=1/4. The evaluation point of the inverse is the output 7, not the input 2.
References
Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen and Christopher J. Stocker, Calculus for Business, Economics, Life Sciences, and Social Sciences, 14th Global Edition, chapters on limits, differentiation and applications. Mark D. Tomenes, Applied Calculus for Business and Economics I, teaching notes and assessment materials (2025–2026).
Further reading: Gilbert Strang and Edwin Herman, Calculus, Volume 1, OpenStax, chapters 2–4.
