Introduction When you need the derivative of two functions multiplied together , you use the product rule .
The Formula If
y = f ( x ) ⋅ g ( x ) y = f(x) \cdot g(x) y = f ( x ) ⋅ g ( x ) , where both
f f f and
g g g are functions of
x x x :
d d x [ f ⋅ g ] = f ′ ⋅ g + f ⋅ g ′ \frac{d}{dx}[f \cdot g] = f' \cdot g + f \cdot g' d x d [ f ⋅ g ] = f ′ ⋅ g + f ⋅ g ′ In words:
Take the derivative of the first function Multiply by the second function unchanged Add the first function unchanged Multiplied by the derivative of the second Memory aid: “first times derivative of second, plus derivative of first times second.”
Alternative Notations Different textbooks use different letters:
d d x [ u ⋅ v ] = u ′ v + u v ′ or d d x [ a ⋅ b ] = a ′ b + a b ′ \frac{d}{dx}[u \cdot v] = u'v + uv' \qquad \text{or} \qquad \frac{d}{dx}[a \cdot b] = a'b + ab' d x d [ u ⋅ v ] = u ′ v + u v ′ or d x d [ a ⋅ b ] = a ′ b + a b ′ They all mean the same thing. Use whichever your teacher prefers.
Basic Examples Example 1: Polynomial Times Trig Function Find the derivative of
f ( x ) = 5 x 2 ⋅ cos x f(x) = 5x^2 \cdot \cos x f ( x ) = 5 x 2 ⋅ cos x .
Identify the two pieces:
First: f = 5 x 2 f = 5x^2 f = 5 x 2 f ′ = 10 x f' = 10x f ′ = 10 x Second: g = cos x g = \cos x g = cos x g ′ = − sin x g' = -\sin x g ′ = − sin x Apply the rule:
f ′ ( x ) = ( 10 x ) ( cos x ) + ( 5 x 2 ) ( − sin x ) = 10 x cos x − 5 x 2 sin x f'(x) = (10x)(\cos x) + (5x^2)(-\sin x) = 10x\cos x - 5x^2\sin x f ′ ( x ) = ( 10 x ) ( cos x ) + ( 5 x 2 ) ( − sin x ) = 10 x cos x − 5 x 2 sin x Example 2: Exponential Times Radical Differentiate
g ( x ) = 3 e x ⋅ x g(x) = 3e^x \cdot \sqrt{x} g ( x ) = 3 e x ⋅ x .
First: 3 e x 3e^x 3 e x 3 e x 3e^x 3 e x Second: x \sqrt{x} x 1 2 x \frac{1}{2\sqrt{x}} 2 x 1 g ′ ( x ) = ( 3 e x ) ( x ) + ( 3 e x ) ( 1 2 x ) = 3 e x x + 3 e x 2 x g'(x) = (3e^x)(\sqrt{x}) + (3e^x)\left(\frac{1}{2\sqrt{x}}\right) = 3e^x\sqrt{x} + \frac{3e^x}{2\sqrt{x}} g ′ ( x ) = ( 3 e x ) ( x ) + ( 3 e x ) ( 2 x 1 ) = 3 e x x + 2 x 3 e x Can you simplify? Yes, factor out
3 e x 3e^x 3 e x :
g ′ ( x ) = 3 e x ( x + 1 2 x ) g'(x) = 3e^x\left(\sqrt{x} + \frac{1}{2\sqrt{x}}\right) g ′ ( x ) = 3 e x ( x + 2 x 1 ) Get a common denominator inside:
= 3 e x ( 2 x 2 x + 1 2 x ) = 3 e x ( 2 x + 1 2 x ) = 3e^x\left(\frac{2x}{2\sqrt{x}} + \frac{1}{2\sqrt{x}}\right) = 3e^x\left(\frac{2x + 1}{2\sqrt{x}}\right) = 3 e x ( 2 x 2 x + 2 x 1 ) = 3 e x ( 2 x 2 x + 1 ) All of these forms are fine:
3 e x x + 3 e x 2 x 3e^x\sqrt{x} + \frac{3e^x}{2\sqrt{x}} 3 e x x + 2 x 3 e x 3 e x ( x + 1 2 x ) 3e^x\left(\sqrt{x} + \frac{1}{2\sqrt{x}}\right) 3 e x ( x + 2 x 1 ) 3 e x ( 2 x + 1 ) 2 x \frac{3e^x(2x + 1)}{2\sqrt{x}} 2 x 3 e x ( 2 x + 1 ) When NOT to Use the Product Rule Don’t use it for constant multiples. Take
h ( x ) = 7 sin x h(x) = 7\sin x h ( x ) = 7 sin x . That’s a constant times a function, not a product of two variable functions.
h ′ ( x ) = 7 cos x h'(x) = 7\cos x h ′ ( x ) = 7 cos x Just use the constant multiple rule.
Do use it for
h ( x ) = x ⋅ sin x h(x) = x \cdot \sin x h ( x ) = x ⋅ sin x , where both factors contain the variable:
h ′ ( x ) = ( 1 ) ( sin x ) + ( x ) ( cos x ) = sin x + x cos x h'(x) = (1)(\sin x) + (x)(\cos x) = \sin x + x\cos x h ′ ( x ) = ( 1 ) ( sin x ) + ( x ) ( cos x ) = sin x + x cos x Rule of thumb: only use the product rule when both factors contain variables.
More Complex Examples Example 3: Rational Function Times Polynomial Find the derivative of
h ( x ) = ( 1 x + 2 ) ( 3 x 2 − 4 ) h(x) = \left(\dfrac{1}{x} + 2\right)(3x^2 - 4) h ( x ) = ( x 1 + 2 ) ( 3 x 2 − 4 ) .
First: f = 1 x + 2 f = \frac{1}{x} + 2 f = x 1 + 2 f ′ = − 1 x 2 f' = -\frac{1}{x^2} f ′ = − x 2 1 Second: g = 3 x 2 − 4 g = 3x^2 - 4 g = 3 x 2 − 4 g ′ = 6 x g' = 6x g ′ = 6 x h ′ ( x ) = ( − 1 x 2 ) ( 3 x 2 − 4 ) + ( 1 x + 2 ) ( 6 x ) h'(x) = \left(-\frac{1}{x^2}\right)(3x^2 - 4) + \left(\frac{1}{x} + 2\right)(6x) h ′ ( x ) = ( − x 2 1 ) ( 3 x 2 − 4 ) + ( x 1 + 2 ) ( 6 x ) Distribute the first term:
= − 3 + 4 x 2 + ( 1 x + 2 ) ( 6 x ) = -3 + \frac{4}{x^2} + \left(\frac{1}{x} + 2\right)(6x) = − 3 + x 2 4 + ( x 1 + 2 ) ( 6 x ) Distribute the second:
= − 3 + 4 x 2 + 6 + 12 x = -3 + \frac{4}{x^2} + 6 + 12x = − 3 + x 2 4 + 6 + 12 x Combine like terms:
h ′ ( x ) = 12 x + 3 + 4 x 2 h'(x) = 12x + 3 + \frac{4}{x^2} h ′ ( x ) = 12 x + 3 + x 2 4 Example 4: Two Trig Functions Differentiate
f ( x ) = sin x ⋅ cos x f(x) = \sin x \cdot \cos x f ( x ) = sin x ⋅ cos x .
First: sin x \sin x sin x cos x \cos x cos x Second: cos x \cos x cos x − sin x -\sin x − sin x f ′ ( x ) = ( cos x ) ( cos x ) + ( sin x ) ( − sin x ) = cos 2 x − sin 2 x f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x) = \cos^2 x - \sin^2 x f ′ ( x ) = ( cos x ) ( cos x ) + ( sin x ) ( − sin x ) = cos 2 x − sin 2 x You
could also write this as
cos ( 2 x ) \cos(2x) cos ( 2 x ) using a trig identity, but you don’t have to.
Using Tables with the Product Rule Sometimes you get function values in a table instead of actual formulas.
Example 5: Table-Based Problem Find
h ′ ( 3 ) h'(3) h ′ ( 3 ) where
h ( x ) = 2 f ( x ) ⋅ g ( x ) h(x) = 2f(x) \cdot g(x) h ( x ) = 2 f ( x ) ⋅ g ( x ) .
Step 1: Set up the product rule. The 2 is a constant multiple, so it factors out front.
h ′ ( x ) = 2 [ f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x ) ] h'(x) = 2[f'(x) \cdot g(x) + f(x) \cdot g'(x)] h ′ ( x ) = 2 [ f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x )] Step 2: Evaluate at
x = 3 x = 3 x = 3 and substitute from the table.
h ′ ( 3 ) = 2 [ ( − 3 ) ( 2 ) + ( 5 ) ( 4 ) ] = 2 [ − 6 + 20 ] = 2 [ 14 ] = 28 h'(3) = 2[(-3)(2) + (5)(4)] = 2[-6 + 20] = 2[14] = 28 h ′ ( 3 ) = 2 [( − 3 ) ( 2 ) + ( 5 ) ( 4 )] = 2 [ − 6 + 20 ] = 2 [ 14 ] = 28 Example 6: Another Table Problem Using the same table, find
k ′ ( 3 ) k'(3) k ′ ( 3 ) where
k ( x ) = 1 2 [ 4 − g ( x ) ] ⋅ f ( x ) k(x) = \frac{1}{2}[4 - g(x)] \cdot f(x) k ( x ) = 2 1 [ 4 − g ( x )] ⋅ f ( x ) .
Step 1: Identify what’s being multiplied.
First: 4 − g ( x ) 4 - g(x) 4 − g ( x ) − g ′ ( x ) -g'(x) − g ′ ( x ) Second: f ( x ) f(x) f ( x ) f ′ ( x ) f'(x) f ′ ( x ) Step 2: Apply the product rule with the
1 2 \frac{1}{2} 2 1 pulled out.
k ′ ( x ) = 1 2 [ ( − g ′ ( x ) ) ⋅ f ( x ) + ( 4 − g ( x ) ) ⋅ f ′ ( x ) ] k'(x) = \frac{1}{2}[(-g'(x)) \cdot f(x) + (4 - g(x)) \cdot f'(x)] k ′ ( x ) = 2 1 [( − g ′ ( x )) ⋅ f ( x ) + ( 4 − g ( x )) ⋅ f ′ ( x )] Step 3: Evaluate at
x = 3 x = 3 x = 3 .
k ′ ( 3 ) = 1 2 [ ( − 4 ) ( 5 ) + ( 4 − 2 ) ( − 3 ) ] = 1 2 [ − 20 − 6 ] = 1 2 [ − 26 ] = − 13 k'(3) = \frac{1}{2}[(-4)(5) + (4 - 2)(-3)] = \frac{1}{2}[-20 - 6] = \frac{1}{2}[-26] = -13 k ′ ( 3 ) = 2 1 [( − 4 ) ( 5 ) + ( 4 − 2 ) ( − 3 )] = 2 1 [ − 20 − 6 ] = 2 1 [ − 26 ] = − 13 Multiple Forms of the Same Answer Different algebra can give you equivalent answers that look totally different. For
g ( x ) = e x ⋅ x g(x) = e^x \cdot \sqrt{x} g ( x ) = e x ⋅ x :
g ′ ( x ) = e x x + e x 2 x = e x ( x + 1 2 x ) = e x ( 2 x + 1 ) 2 x g'(x) = e^x\sqrt{x} + \frac{e^x}{2\sqrt{x}} = e^x\left(\sqrt{x} + \frac{1}{2\sqrt{x}}\right) = \frac{e^x(2x + 1)}{2\sqrt{x}} g ′ ( x ) = e x x + 2 x e x = e x ( x + 2 x 1 ) = 2 x e x ( 2 x + 1 ) All correct. On free response, any of them should get full credit. On multiple choice, you may have to manipulate yours to match an option.
Common Mistakes Mistake 1: Forgetting to add both terms.
Wrong:
d d x [ x 2 sin x ] = 2 x cos x \frac{d}{dx}[x^2 \sin x] = 2x \cos x d x d [ x 2 sin x ] = 2 x cos x Right:
d d x [ x 2 sin x ] = 2 x sin x + x 2 cos x \frac{d}{dx}[x^2 \sin x] = 2x \sin x + x^2 \cos x d x d [ x 2 sin x ] = 2 x sin x + x 2 cos x You need both terms.
Mistake 2: Using the product rule on constants.
Wrong (well, unnecessary): for
f ( x ) = 5 x 3 f(x) = 5x^3 f ( x ) = 5 x 3 , writing
f ′ ( x ) = 0 ⋅ x 3 + 5 ⋅ 3 x 2 = 15 x 2 f'(x) = 0 \cdot x^3 + 5 \cdot 3x^2 = 15x^2 f ′ ( x ) = 0 ⋅ x 3 + 5 ⋅ 3 x 2 = 15 x 2 Right: just use the constant multiple rule and get
f ′ ( x ) = 15 x 2 f'(x) = 15x^2 f ′ ( x ) = 15 x 2 .
Mistake 3: Mixing up which function is first.
Order doesn’t actually change the final answer since addition is commutative, but staying consistent helps you avoid errors, especially once you get to the quotient rule.
Practice Problems Problem 1. Find the derivative:
f ( x ) = x 3 ⋅ e x f(x) = x^3 \cdot e^x f ( x ) = x 3 ⋅ e x f ′ ( x ) = 3 x 2 e x + x 3 e x = e x ( 3 x 2 + x 3 ) f'(x) = 3x^2 e^x + x^3 e^x = e^x(3x^2 + x^3) f ′ ( x ) = 3 x 2 e x + x 3 e x = e x ( 3 x 2 + x 3 ) Problem 2. Differentiate:
g ( x ) = ( 2 x + 1 ) ( sin x ) g(x) = (2x + 1)(\sin x) g ( x ) = ( 2 x + 1 ) ( sin x ) g ′ ( x ) = 2 sin x + ( 2 x + 1 ) cos x g'(x) = 2\sin x + (2x + 1)\cos x g ′ ( x ) = 2 sin x + ( 2 x + 1 ) cos x Problem 3. Find
h ′ ( x ) h'(x) h ′ ( x ) for
h ( x ) = x ln x h(x) = x \ln x h ( x ) = x ln x h ′ ( x ) = ( 1 ) ( ln x ) + ( x ) ( 1 x ) = ln x + 1 h'(x) = (1)(\ln x) + (x)\left(\frac{1}{x}\right) = \ln x + 1 h ′ ( x ) = ( 1 ) ( ln x ) + ( x ) ( x 1 ) = ln x + 1 Problem 4. Find
p ′ ( 2 ) p'(2) p ′ ( 2 ) where
p ( x ) = f ( x ) ⋅ g ( x ) p(x) = f(x) \cdot g(x) p ( x ) = f ( x ) ⋅ g ( x ) :
p ′ ( 2 ) = ( 4 ) ( − 1 ) + ( 3 ) ( 5 ) = − 4 + 15 = 11 p'(2) = (4)(-1) + (3)(5) = -4 + 15 = 11 p ′ ( 2 ) = ( 4 ) ( − 1 ) + ( 3 ) ( 5 ) = − 4 + 15 = 11 Key Takeaways Product rule: d d x [ f ⋅ g ] = f ′ ⋅ g + f ⋅ g ′ \frac{d}{dx}[f \cdot g] = f' \cdot g + f \cdot g' d x d [ f ⋅ g ] = f ′ ⋅ g + f ⋅ g ′ Use it when both factors contain variables Don’t use it for constant multiples Both terms have to be there, it’s a sum Order doesn’t affect the final answer Multiple simplified forms can all be correct On multiple choice you may need to manipulate your answer to match With tables, read the correct row carefully Factor out constants at the start to make life easier 
