Finding a derivative from the definition

When you reach for it

The derivative at a point is one limit for one number: pick a, take the limit of the secant slopes, get the slope of the curve at a. Ask about a second point and the whole limit is done again.

Doing the limit once with x left as a letter gives every slope at once. The result is a new function, the derivative f'(x), whose value at each input is the slope of the graph of f there. Reach for the definition when you have a rule and no differentiation rules yet, when you want to see where a rule comes from, or when a function is new enough that no rule covers it.

The raw material is the difference quotient \displaystyle \frac{f(x + h) - f(x)}{h}, the slope of the secant from x to x + h, which you already know how to simplify. The one new step is the limit.

The procedure

The derivative of f is

at every x where this limit exists. In Leibniz notation the same function is written \dfrac{d}{dx}\big[f(x)\big], or \dfrac{dy}{dx} when y = f(x). Inside the limit x is held fixed and only h moves, so x is treated like a constant throughout. Evaluating the limit by hand takes three moves.

Simplify the quotient for h \ne 0. Substituting h = 0 straight away gives \displaystyle \frac{f(x) - f(x)}{0} = \frac{0}{0}, every time, because the two points have merged. As in limits by algebra, \displaystyle \frac{0}{0} says a factor of h is hiding in the numerator. The shape of f says how to expose it:

• a polynomial: expand f(x + h); the terms without h cancel against f(x) and every surviving term carries h;
• a square root: rationalize, multiplying top and bottom by the conjugate so that (A - B)(A + B) = A^2 - B^2 clears the roots from the numerator;
• a fraction: combine f(x + h) - f(x) over a common denominator; the numerator collapses to a multiple of h.

Then cancel the h, which is allowed because h \ne 0.

Let h go to 0. A limit as h \to 0 looks only at h \ne 0, and for every such h the simplified form equals the original quotient, so the two have the same limit. The simplified form no longer divides by h; it is continuous at h = 0 (a polynomial, or a quotient of continuous pieces whose bottom is not zero there), so its limit is its value at h = 0. Substitution is allowed now and was not before.

Check against a point. Pick an input a and compare f'(a) with the quotient for a small h, computed straight from f. The two should agree to about as many places as h is small. The check catches a dropped sign, and it is the bridge back to the slope of the graph at one point.

The limit may fail to exist at some inputs, so the domain of f' can be smaller than the domain of f. Say which x the answer holds for.

Worked examples

Boundaries

• The limit has to exist. f'(a) is defined only where \lim_{h \to 0} of the quotient is a finite number. It fails at a corner such as |x| at 0, where the two one-sided limits differ, and at a vertical tangent such as \displaystyle \sqrt{2x + 1} at \displaystyle -\frac{1}{2}, where the quotient grows without bound: f is defined there and f' is not.
• x must be in the domain of f, with room on both sides. The quotient needs f(x) and f(x + h) for all small h of both signs. So \displaystyle \frac{3}{x + 2} has no derivative at -2, and the domain of f' is often smaller than the domain of f.
• Substitute only after the h in the denominator is gone. Cancelling h uses h \ne 0, which the limit allows. Putting in h = 0 is allowed only once what is left is continuous at h = 0, with a bottom that is not 0 there.

Where this goes next

Every derivative here took a page of algebra, and the answers already hint at shortcuts: 2x, 3x^2, and in the exercises 4x^3. The power rule and basic differentiation rules turn the hint into a theorem, proved from this same limit. The product rule is found the same way, by evaluating the limit for a product. The derivative as a function reads f' off a graph when no formula is given, and the equation of a tangent line puts f'(a) to work.

Exercises

Expanding

Find f'(x) from the definition, showing the simplified difference quotient.

1. f(x) = 3x^2 - 7x
2. g(x) = 5 - x^2
3. p(x) = 4x - 9
4. q(x) = x^3 + 2x
5. r(x) = x^4
6. k(x) = 2x^3 - x^2

Rationalizing

Find the derivative from the definition, and say for which x it holds.

7. \displaystyle f(x) = \sqrt{x}
8. \displaystyle g(x) = \sqrt{5 - x}
9. \displaystyle s(x) = \sqrt{3x - 2}

Combining fractions

Find the derivative from the definition, and say for which x it holds.

10. f(x) = \dfrac{1}{x - 4}
11. g(x) = \dfrac{6}{1 - 3x}
12. w(x) = \dfrac{x}{x - 1}
13. \displaystyle u(x) = \dfrac{1}{\sqrt{x}} (combine, then rationalize)

Checking at a point

14. Use exercise 1 to find f'(2) for f(x) = 3x^2 - 7x. Then find the secant slopes with h = 0.01 and h = -0.01 directly from f.
15. Use exercise 7 to find the slope of \displaystyle y = \sqrt{x} at x = 9, and compare it with the secant slope for h = 0.01.
16. Use exercise 4 to find every x at which y = x^3 + 2x has slope 14.

Ziga on the track

Ziga coasts after a sprint, covering d(t) = 15t - 0.2t^2 meters in the first t seconds, for 0 \le t \le 10.

17. Find d'(t) from the definition, and his speed at t = 5, with units.
18. Find his average speeds on [4, 5] and on [5, 6], and compare them with d'(5).
19. At what time has his speed dropped to 12 meters per second?

Reasoning

20. Show from the definition that f(x) = ax^2 + bx + c has f'(x) = 2ax + b. Check it against Example 1 and exercise 1.
21. Nika puts h = 0 into \displaystyle \frac{(x + h)^3 - x^3}{h}, gets \displaystyle \frac{0}{0}, and concludes that x^3 has no derivative. Explain what went wrong.
22. Show from the definition that f(x) = |x^2 - 4| has no derivative at x = 2, by finding both one-sided limits of the quotient.
23. Explain why \displaystyle g(x) = \sqrt{5 - x} from exercise 8 has no derivative at x = 5 although g(5) is defined.
24. In exercise 9, Marek asks why multiplying top and bottom by the conjugate is allowed. Explain, and say which x and h the step needs.