Product and Quotient Rules: Differentiating Multiplication and Division

Why “differentiate each factor” fails

You already know how to differentiate sums and constant multiples term-by-term. A common mistake is to extend that idea to multiplication and division, but it does not work in general.

Counterexample for products

Let u=x and v=x. Then uv=x^2.

• Incorrect: u'v' = 1·1 = 1
• Correct: (x^2)' = 2x

Since 1 is not equal to 2x (for most x), the “multiply the derivatives” rule is false.

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Counterexample for quotients

Let u=x and v=x^2. Then u/v = x/x^2 = 1/x.

• Incorrect: u'/v' = 1/(2x)
• Correct: (1/x)' = -1/x^2

Again, these are not the same function, so “divide the derivatives” is false.

The product rule (multiplication)

If y=u(x)v(x), then

(uv)' = u'v + uv'

Step template (product rule)

• 1) Identify u and v (the two factors).
• 2) Compute u' and v'.
• 3) Apply (uv)' = u'v + uv'.
• 4) Simplify (factor, combine like terms, rewrite with positive exponents if desired).

Worked example 1: polynomial times polynomial

Differentiate y=(x^2+3x)(2x-5).

• Choose u=x^2+3x, v=2x-5
• Then u'=2x+3, v'=2

Apply the rule:

y' = u'v + uv' = (2x+3)(2x-5) + (x^2+3x)(2)

Simplify (one clean route is expand then combine):

(2x+3)(2x-5)=4x^2-10x+6x-15=4x^2-4x-15

2(x^2+3x)=2x^2+6x

So

y' = (4x^2-4x-15) + (2x^2+6x) = 6x^2+2x-15

Worked example 2: radicals rewritten as exponents

Differentiate y=(x^3-1)\sqrt{x}.

Rewrite \sqrt{x}=x^{1/2}, so y=(x^3-1)x^{1/2}.

• u=x^3-1 so u'=3x^2
• v=x^{1/2} so v'=(1/2)x^{-1/2}

Apply product rule:

y' = 3x^2·x^{1/2} + (x^3-1)·(1/2)x^{-1/2}

Simplify strategically by combining exponents:

• 3x^2·x^{1/2} = 3x^{5/2}
• Factor out x^{-1/2} if you want a single power outside:

y' = x^{-1/2}\left(3x^3 + \tfrac12(x^3-1)\right) = x^{-1/2}\left(\tfrac72 x^3 - \tfrac12\right)

Optional rewrite without negative exponents:

y' = \dfrac{\tfrac72 x^3 - \tfrac12}{\sqrt{x}} = \dfrac{7x^3-1}{2\sqrt{x}}

The quotient rule (division)

If y=u(x)/v(x) with v(x)\neq 0, then

(u/v)' = (u'v - uv')/v^2

Notice the order: u'v minus uv', all over v^2.

Step template (quotient rule)

• 1) Identify u (top) and v (bottom).
• 2) Compute u' and v'.
• 3) Apply (u/v)' = (u'v - uv')/v^2.
• 4) Simplify (combine terms in the numerator, factor if helpful; keep denominator as v^2 unless there is a clear cancellation).

Worked example 3: rational function (polynomial over polynomial)

Differentiate y=(x^2+1)/(x-2).

• u=x^2+1 so u'=2x
• v=x-2 so v'=1

Apply quotient rule:

y' = \dfrac{(2x)(x-2) - (x^2+1)(1)}{(x-2)^2}

Simplify numerator:

(2x)(x-2)=2x^2-4x

2x^2-4x-(x^2+1)=x^2-4x-1

So

y' = \dfrac{x^2-4x-1}{(x-2)^2}

Worked example 4: quotient with radicals rewritten as exponents

Differentiate y=\dfrac{\sqrt{x}+1}{x^2}.

Rewrite \sqrt{x}=x^{1/2}. Choose u=x^{1/2}+1, v=x^2.

• u'=(1/2)x^{-1/2}
• v'=2x

Apply quotient rule:

y' = \dfrac{\left((1/2)x^{-1/2}\right)(x^2) - (x^{1/2}+1)(2x)}{(x^2)^2}

Simplify step-by-step:

• ((1/2)x^{-1/2})(x^2)=(1/2)x^{3/2}
• (x^{1/2}+1)(2x)=2x^{3/2}+2x

So numerator is

(1/2)x^{3/2} - (2x^{3/2}+2x) = -(3/2)x^{3/2} - 2x

Denominator: (x^2)^2=x^4. Therefore

y' = \dfrac{-(3/2)x^{3/2} - 2x}{x^4}

Rewrite with negative exponents (often cleaner):

y' = -(3/2)x^{-5/2} - 2x^{-3}

Side-by-side: incorrect vs correct on the same function

Example A (product)

Differentiate y=(x^2+1)(x^3+1).

The results disagree (for example at x=1: incorrect gives 6, correct gives 10).

Example B (quotient)

Differentiate y=\dfrac{x^2+1}{x}.

At x=2, incorrect gives 4, correct gives (4-1)/4=3/4.

Simplify strategically: before vs after differentiating

When simplifying before differentiating helps

If you can rewrite a quotient as a simpler expression using algebraic cancellation or splitting, you may avoid the quotient rule entirely.

Example: y=\dfrac{x^2+1}{x} = x + x^{-1} (valid for x\neq 0).

Then

y' = 1 - x^{-2} = 1 - \dfrac{1}{x^2}

This is usually faster and less error-prone than applying the quotient rule.

When simplifying before differentiating is dangerous

Expanding products can create more terms than necessary, increasing arithmetic mistakes.

Example: y=(x^2+3x)(2x-5)

• If you expand first, you must expand carefully, then differentiate a longer polynomial.
• If you use the product rule first, you keep structure and often simplify more reliably.

When simplifying after differentiating is safer

With quotients, canceling factors before differentiating can hide domain restrictions or lead you to cancel something that is not a common factor (only common factors can cancel, not common terms).

Example of a common pitfall: \dfrac{x^2-1}{x-1} can simplify because x^2-1=(x-1)(x+1), but you must factor first. You cannot “cancel” the -1 with the -1 in the denominator unless it is a factor.

A safe workflow is: factor carefully, cancel only common factors, note excluded values (here x\neq 1), then differentiate the simplified form if it truly is simpler.

Mixed practice: choose the correct rule (with brief justification)

For each function, (1) state which rule(s) you will use (power vs product vs quotient), (2) give a one-sentence justification, and (3) compute y'.

• y = (3x^2-4x+1)(x^5) (justify product vs expanding)
• y = \dfrac{x^3-2x}{x^2+1} (justify quotient)
• y = \dfrac{x^2+1}{x} (justify simplify first vs quotient rule)
• y = (x^{1/2}+x^{-1/2})(x^2-1) (justify rewriting radicals as exponents, then product)
• y = \dfrac{(x-1)(x+2)}{x-1} (justify simplify first, note domain restriction, then power/linearity)
• y = \dfrac{\sqrt{x}}{x^3} (justify rewriting as a single power vs quotient rule)
• y = (x^2+1)^2 (decide: is this product of identical factors or something else? justify your choice for this chapter’s tools)
• y = \dfrac{2x-5}{\sqrt{x}} (justify rewriting as exponents and choosing quotient vs product with a negative exponent)