2026 Fall Section 3.4 The Product & Quotient Rules
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Overview
Ara Araeipour reviews derivative definitions and basic formulas, then develops the product rule, (fg)' = f'g + g'f, and the quotient rule, (f/g)' = (f'g - g'f)/g². Examples apply the rules to polynomial, exponential, and trigonometric functions, distinguish calculus steps from algebraic simplification, and use derivatives to find tangent lines.
Key takeaways
- The product rule is (fg)' = f'g + g'f, not f'g'; the example x·x² gives derivative 3x², while multiplying the individual derivatives gives only 2x.
- The quotient rule is (f/g)' = (f'g - fg')/g²; keeping the numerator order and squaring the entire denominator prevents common setup errors.
- For √x e^x, rewriting √x as x^(1/2) makes the product rule straightforward and yields e^x(2x + 1)/(2√x).
- Before using the quotient rule, check whether algebraic rewriting makes the power rule simpler: (t - √t)/t^(1/3) becomes t^(2/3) - t^(1/6).
- A tangent-line equation requires both the point and the derivative evaluated at its x-coordinate; for x/(2x - 7) at x = 3, the slope is -7 and the line is y = -7x + 18.
- Trigonometric identities can finish derivative simplifications: sin²x + cos²x = 1 proves (tan x)' = sec²x, and sec²x - tan²x = 1 simplifies the derivative of (1 - sec x)/tan x.
Chapters
- Connect the secant-line slope, rise over run, to the tangent-line slope and instantaneous rate of change.
- Review derivative definitions using limits with x approaching a and h approaching zero.
- Recall constant, power, exponential, and trigonometric derivatives, including (x^n)' = nx^(n-1) and (e^x)' = e^x.
- State the product rule as (fg)' = f'g + g'f, rather than multiplying the derivatives.
- Disprove the tempting but incorrect formula with f(x) = x and g(x) = x²: (x³)' = 3x², while f'g' = 2x.
- Verify the correct rule for the same example: 1·x² + 2x·x = 3x².
- Derive the product rule from the limit definition by adding and subtracting f(x+h)g(x), then applying the limit laws.
- Differentiate a product of polynomial factors by identifying f and g first; Araeipour emphasizes that applying the rule is the calculus step and expansion is algebra.
- For h(x) = xe^x, obtain h'(x) = e^x(1+x), then differentiate again to get h''(x) = e^x(x+2).
- For y = x² sin x, apply the rule to get y' = 2x sin x + x² cos x.
- Set F(x) = f(x)/g(x) and substitute it into the derivative limit definition.
- Clear the complex fraction by multiplying by g(x)g(x+h), then add and subtract f(x)g(x) in the numerator.
- Apply the limits to derive (f/g)' = (f'g - fg')/g², with the denominator squared.
- Set up the quotient-rule derivative for a rational function with numerator x² - 3x and denominator x - 1, then simplify its numerator to x² - 2x + 3.
- Use the discriminant, b² - 4ac = -8, to show x² - 2x + 3 does not factor over the reals and cannot cancel the denominator factor x - 1.
- For y = (x² - 3x)/(x - 1), keep the derivative as (x² - 2x + 3)/(x - 1)² after checking for cancellation.
- Rewrite √x as x^(1/2) before differentiating √x e^x; combining terms gives h'(x) = e^x(2x + 1)/(2√x).
- For e^x/(1+x), the quotient rule simplifies to xe^x/(1+x)².
- Differentiate (1 + sin x)/(x + cos x) with the quotient rule; using 1 - sin²x = cos²x reduces the numerator to x cos x.
- Rewrite (t - √t)/t^(1/3) as t^(2/3) - t^(1/6) to use the power rule instead of the quotient rule.
- Write tan x as sin x/cos x and apply the quotient rule; the identity sin²x + cos²x = 1 yields (tan x)' = sec²x.
- Write sec x as 1/cos x and use the quotient rule to obtain (sec x)' = sec x tan x.
- Araeipour recommends using the same reciprocal-function and quotient-rule approach to derive the derivatives of cot x and csc x.
- For y = x/(2x - 7), differentiate to get y' = -7/(2x - 7)²; at x = 3, the point is (3, -3) and the slope is -7.
- Use point-slope form at (3, -3) to obtain the tangent line y = -7x + 18.
- For h(m) = (9m - 7)/(5m + 4), apply the quotient rule and simplify the numerator to 71, giving h'(m) = 71/(5m + 4)².
- Apply the product rule to y = e^x cos x, giving y' = e^x(cos x - sin x).
- Evaluate the derivative at x = 0 using e⁰ = 1, cos 0 = 1, and sin 0 = 0 to find slope 1.
- Combine slope 1 with the point (0, 1) to obtain the tangent line y = x + 1.
- Differentiate y = (1 - sec x)/tan x using the quotient rule, with (sec x)' = sec x tan x and (tan x)' = sec²x.
- Simplify the numerator using sec²x - tan²x = 1.
- Write the final derivative as y' = sec x(1 - sec x)/tan²x.
Transcript excerpt
0:07 Good morning class. Here's chapter 3, derivatives, section 3.4, the product and quotient rules. We've covered the derivatives and rates of change, derivative as a function, basic differentiation formulas, and now we're going to move on to the product and quotient rules. Reminding you of various differentiation formulas starting with a secant line. When we have two points, we can calculate the slope known as rise over run. differentiating it. If you if you move
0:54 to the red line becomes the tangent line and it becomes the instantaneous rate of change at a and the equation of a tangent. So frime of a becomes limit of f of x - f of ax - a as x approaches a or limit of f of a plus h - f of a h as h approaches zero. And here's the point slope that gives rise to the equation of a red line, the tangent line. Then going back to the definition, dy dx is limit of delta y where delta x delta x approaches zero. frime of x is limit of f of x plus h minus f of x over h as h approaches zero. This one is the liability notation. The derivative of x is one. If you have the derivative of c
1:41 f of x c comes out and f of x plus minus g of x you just do them term by term when you want to these are various formulas we should be very comfortable with the derivative of the constant is zero the derivative of kx is k and this is very important the power the derivative of x the^ n is n x^ nus one e the^ of x the derivative is itself the derivative of s is cosine the dative of cosine is negative s. The derivative of
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Professor Ara Araeipour.