Limits and Derivatives · Lesson 5 of 8
Derivatives from First Principles
“Turn the limiting rate idea into a definition and calculate derivatives step by step.”
• Construct a difference quotient at a point and at a variable input. • Calculate derivatives from first principles using limit laws. • Interpret a derivative as a local rate and a tangent slope. • Distinguish f(a), f′(a), and the derivative function f′(x). • Identify where a derivative expression is defined and where its limit fails.
A precise definition of instantaneous change
We estimated the velocity of a falling body by comparing a fixed time with nearby times. Now replace distance by a general output f(x), and replace time by a general input x. This makes the same reasoning useful whenever one quantity depends on another.
A reservoir manager may compare water depths at nearby times to understand how quickly the level is rising. A rocket scientist may compare positions to determine velocity. A financial model may describe how a value changes with time. Calculating a derivative describes the local rate of a given function; predictions also depend on whether that function is an appropriate model.
Fix a point a in the domain of f. Move the input by a small increment h, so the new input is a+h. The output change is f(a+h)−f(a). Dividing by h gives the average rate over that small change. The derivative is the limit of that rate as h approaches zero, provided the required finite limit exists.
The derivative of f at a, written f′(a), is the limit of [f(a+h)-f(a)]/h as non-zero h approaches zero, if the limit exists as a finite real number.
The input a stays fixed throughout the calculation; h varies. The numerator is a difference of two outputs, not a formula to differentiate separately. We first simplify the quotient for h≠0 and only then take its limit. Both f(a) and the nearby values f(a+h) must be meaningful for the approach we use.
Do not set h=0 in the original difference quotient. That would produce 0/0. A derivative is found by a limiting process over non-zero increments.
Derivatives at particular inputs
For a linear function, equal input changes produce equal output changes. We should therefore expect the same rate everywhere. A quadratic behaves differently: its local rate changes with the position at which we take the derivative. The next calculations let the definition reveal this contrast.
Problem
For f(x)=3x, find f′(2) from first principles.
- 1.The point value is f(2)=6. The nearby value is f(2+h)=3(2+h)=6+3h.
- 2.Subtract to get f(2+h)−f(2)=3h, then divide by the non-zero increment h.
- 3.The quotient is 3 for every allowed h≠0, so its limit is 3. Thus f′(2)=3.
Problem
Let f(x)=2x²+3x−5. Find f′(−1) and f′(0), and verify f′(0)+3f′(−1)=0.
- 1.At a=−1, f(−1)=−6. Expand f(−1+h)=2(1−2h+h²)+3(−1+h)−5=−6−h+2h².
- 2.The output change is −h+2h². For h≠0, the difference quotient is −1+2h, whose limit is −1. Therefore f′(−1)=−1.
- 3.At a=0, f(0)=−5 and f(h)=2h²+3h−5. The difference quotient is (2h²+3h)/h=2h+3.
- 4.Its limit is 3, so f′(0)=3. The requested relation is 3+3(−1)=0.
- 5.The negative derivative at −1 means a locally decreasing output; the positive derivative at 0 means a locally increasing output.
Problem
Find the derivative of f(x)=sin x at x=0, using radians.
- 1.Since sin 0=0, the difference quotient is [sin(0+h)−0]/h=sin h/h.
- 2.The standard trigonometric limit gives limit 1.
- 3.Thus f′(0)=1. This calculation shows why the earlier sine limit is needed for derivatives.
Problem
For f(x)=3, find f′(0) and f′(3).
- 1.At either input, both the original and nearby output equal 3.
- 2.The numerator of the difference quotient is 3−3=0, and 0/h=0 for every non-zero h.
- 3.Both derivatives are zero. The function value is 3 but the local rate of change is 0.
The tangent interpretation
Let P=(a,f(a)) and Q=(a+h,f(a+h)) on the graph. The horizontal displacement from P to Q is h and the vertical displacement is f(a+h)−f(a). Their ratio is the slope of the secant PQ. This remains valid for negative h as well as positive h, because both signed changes are used.
When Q moves towards P, the secant approaches the tangent if the derivative exists. The derivative is therefore the tangent slope. If the tangent makes an angle ψ with the positive horizontal direction and has a finite slope, f′(a)=tan ψ. This connects a rate calculation to a visible geometric quantity.
The derivative is a slope, not the tangent line itself. To name a tangent line fully we would also need its point of contact. Here the chapter concentrates on finding and interpreting the slope.
A derivative can be a new function
Instead of repeating the calculation at a=1, a=2, and every other input, keep the input as the variable x. A formula for f′(x) supplies the derivative at every point where the defining limit exists. We can then substitute a into that new formula to obtain f′(a).
This is called differentiation from first principles because it returns to the definition. In the calculation, x is treated as fixed while h approaches zero. Only after the limit is evaluated do we view the answer as a function of x.
| Notation | Meaning |
|---|---|
| f′(x) | Derivative function evaluated at the input x |
| df/dx or d/dx[f(x)] | Derivative with respect to x |
| dy/dx when y=f(x) | Rate of change of y with respect to x |
| D(f(x)) | Another notation for the same derivative |
| f′(a) or (df/dx) at x=a | The derivative value at the specified point a |
Problem
Find the derivative of f(x)=10x from first principles.
- 1.Write f(x+h)=10x+10h and f(x)=10x.
- 2.The difference quotient is [(10x+10h)−10x]/h=10h/h=10 for h≠0.
- 3.The limit is 10, so f′(x)=10 at every real x.
Problem
Find the derivative of f(x)=x² from first principles.
- 1.Expand f(x+h)=(x+h)²=x²+2xh+h².
- 2.Subtract x² and divide by h≠0: (2xh+h²)/h=2x+h.
- 3.As h→0 with x held fixed, 2x+h→2x. Therefore f′(x)=2x.
- 4.At x=1 the derivative is 2, and at x=3 it is 6. The derivative function captures how the tangent slope changes along the curve.
Problem
For a fixed real constant c, differentiate f(x)=c.
- 1.Both f(x+h) and f(x) equal c.
- 2.The difference quotient is (c−c)/h=0 for h≠0.
- 3.Thus f′(x)=0. The symbol c stands for a fixed number, not a varying input.
Problem
Differentiate f(x)=1/x and state its domain.
- 1.The function requires x≠0. Choose h close enough to zero that x+h is also non-zero.
- 2.The quotient is [1/(x+h)-1/x]/h. Combining the numerator gives [-h/(x(x+h))]/h.
- 3.Cancel h for h≠0 to obtain -1/[x(x+h)]. Its limit is -1/x².
- 4.Therefore f'(x)=-1/x² for x≠0. Neither the function nor this derivative is defined at zero.
Existence is part of the definition
A formula for a difference quotient is only a candidate for a derivative. Its limit must exist. A corner in a graph can make the slopes from the two sides disagree even when the function itself has a limit and a value at the point. This is the same one-sided reasoning used earlier, now applied to the rate quotient.
Problem
Use the definition to test whether f(x)=|x| has a derivative at zero.
- 1.The difference quotient is |h|/h because f(0)=0.
- 2.For h>0 the quotient is 1; for h<0 it is −1.
- 3.Its two side limits disagree, so no finite two-sided derivative exists at zero. The function limit is nevertheless zero.
- 4.This example is an illustration of the existence condition in the definition, not a new differentiation rule.
Quiz
What is f′(2) when f(x)=3x?
For f(x)=x², what is f′(3)?
What remains fixed while taking the first-principles limit at a?
If f(x)=7 everywhere, what is f′(x)?
What is the derivative of 1/x where it is defined?
Which statement describes f′(a)?
Practice Problems
- Use first principles to find the derivative of f(x)=x²−2 at x=10.
- Use first principles to find the derivative of f(x)=x at x=1.
- Use first principles to find the derivative of f(x)=99x at x=100.
- From first principles, differentiate each of x³-27, (x-1)(x-2), 1/x², and (x+1)/(x-1). State all excluded inputs.
- For f(x)=2x²+3x−5, find f′(x) from first principles and use it to check the derivative values at −1 and 0.
- For s(t)=4.9t², calculate s′(t) from first principles and explain its units. Check the velocity at t=2.
- Explain why f(a)=0 does not imply f′(a)=0. Give a linear-function example.
- At the corner of y=|x|, compare the left and right difference-quotient limits at zero. Explain why the derivative fails although the function limit exists.
Key Takeaways
• A derivative is the finite limit of a difference quotient as its non-zero increment approaches zero. • The derivative at a point is a number; the derivative function gives those numbers across its domain. • The difference quotient is a secant slope, and its limiting value is the tangent slope. • Constant functions have derivative zero; changing functions can have input-dependent derivatives. • Domain restrictions and the existence of the limiting rate remain essential.