Lecture 21: Multiplication and Division Rules
Watch 3Blue1Brown videos on Derivative formulas through geometry and visualizing the chain rule and product rule for a further conceptual understanding of the rules.
Constant Multiple Rule
If f ( x ) f(x) f ( x ) is a differentiable function and c c c is a constant, then:
d d x [ c ⋅ f ( x ) ] = c ⋅ d d x [ f ( x ) ] \frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]
d x d [ c ⋅ f ( x )] = c ⋅ d x d [ f ( x )]
Example:
d d x [ 7 x 2 ] = 7 ⋅ d d x [ x 2 ] = 7 ⋅ 2 x = 14 x \frac{d}{dx}[7x^2] = 7 \cdot \frac{d}{dx}[x^2] = 7 \cdot 2x = 14x d x d [ 7 x 2 ] = 7 ⋅ d x d [ x 2 ] = 7 ⋅ 2 x = 14 x
This rule lets us factor out constants when taking derivatives.
Multiplication Rule (Product Rule)
If f ( x ) f(x) f ( x ) and g ( x ) g(x) g ( x ) are both differentiable, then:
d d x [ f ( x ) ⋅ g ( x ) ] = f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x ) \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)
d x d [ f ( x ) ⋅ g ( x )] = f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x )
“First times derivative of second, plus second times derivative of first.”
Example 1:
f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 , g ( x ) = x + 1 g(x) = x+1 g ( x ) = x + 1
Then:
d d x [ ( x 2 ) ( x + 1 ) ] = 2 x ( x + 1 ) + x 2 ( 1 ) = 2 x 2 + 2 x + x 2 = 3 x 2 + 2 x \frac{d}{dx}[(x^2)(x+1)] = 2x(x+1) + x^2(1) = 2x^2 + 2x + x^2 = 3x^2 + 2x
d x d [( x 2 ) ( x + 1 )] = 2 x ( x + 1 ) + x 2 ( 1 ) = 2 x 2 + 2 x + x 2 = 3 x 2 + 2 x
Example 2:
d d x [ ( x 2 + 1 ) ( x 3 − 2 x ) ] \frac{d}{dx}[(x^2+1)(x^3-2x)] d x d [( x 2 + 1 ) ( x 3 − 2 x )]
Apply the product rule:
First: f ( x ) = x 2 + 1 f(x) = x^2+1 f ( x ) = x 2 + 1 , so f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x
Second: g ( x ) = x 3 − 2 x g(x) = x^3 - 2x g ( x ) = x 3 − 2 x , so g ′ ( x ) = 3 x 2 − 2 g'(x) = 3x^2 - 2 g ′ ( x ) = 3 x 2 − 2
So:
d d x [ ( x 2 + 1 ) ( x 3 − 2 x ) ] = 2 x ( x 3 − 2 x ) + ( x 2 + 1 ) ( 3 x 2 − 2 ) = 2 x 4 − 4 x 2 + 3 x 2 − 2 = \frac{d}{dx}[(x^2+1)(x^3-2x)] = 2x(x^3 - 2x) + (x^2+1)(3x^2 - 2) = 2x^4 - 4x^2 + 3x^2 - 2 =
d x d [( x 2 + 1 ) ( x 3 − 2 x )] = 2 x ( x 3 − 2 x ) + ( x 2 + 1 ) ( 3 x 2 − 2 ) = 2 x 4 − 4 x 2 + 3 x 2 − 2 =
= 2 x 4 − x 2 − 2 = 2x^4 - x^2 - 2
= 2 x 4 − x 2 − 2
Division Rule (Quotient Rule)
If f ( x ) f(x) f ( x ) and g ( x ) g(x) g ( x ) are differentiable, and g ( x ) ≠ 0 g(x) \ne 0 g ( x ) = 0 , then:
d d x [ f ( x ) g ( x ) ] = f ′ ( x ) ⋅ g ( x ) − f ( x ) ⋅ g ′ ( x ) [ g ( x ) ] 2 \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}
d x d [ g ( x ) f ( x ) ] = [ g ( x ) ] 2 f ′ ( x ) ⋅ g ( x ) − f ( x ) ⋅ g ′ ( x )
“Low d-high minus high d-low over square of what’s below.”
Example 1:
d d x [ x 2 x + 1 ] \frac{d}{dx}\left[\frac{x^2}{x+1}\right] d x d [ x + 1 x 2 ]
Let f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 , g ( x ) = x + 1 g(x) = x+1 g ( x ) = x + 1
Then:
2 x ( x + 1 ) − x 2 ( 1 ) ( x + 1 ) 2 = 2 x ( x + 1 ) − x 2 ( x + 1 ) 2 \frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{2x(x+1) - x^2}{(x+1)^2}
( x + 1 ) 2 2 x ( x + 1 ) − x 2 ( 1 ) = ( x + 1 ) 2 2 x ( x + 1 ) − x 2
Example 2:
d d x [ 1 x 2 ] \frac{d}{dx} \left[\frac{1}{x^2}\right] d x d [ x 2 1 ]
Use quotient rule:
0 ⋅ x 2 − 1 ⋅ 2 x x 4 = − 2 x x 4 = − 2 x 3 \frac{0\cdot x^2 - 1\cdot 2x}{x^4} = \frac{-2x}{x^4} = -\frac{2}{x^3}
x 4 0 ⋅ x 2 − 1 ⋅ 2 x = x 4 − 2 x = − x 3 2
which will be the same answer if you just apply the power rule to x − 2 x^{-2} x − 2 :
d d x [ x − 2 ] = − 2 x − 2 − 1 = − 2 x − 3 = − 2 x 3 \frac{d}{dx}[x^{-2}] = -2x^{-2 - 1} = -2x^{-3} = -\frac{2}{x^3}
d x d [ x − 2 ] = − 2 x − 2 − 1 = − 2 x − 3 = − x 3 2
With this rule, we can differentiate any rational function. For example,
d d x [ x 2 + 1 x 2 − 1 ] = ( 2 x ) ( x 2 − 1 ) − ( x 2 + 1 ) ( 2 x ) ( x 2 − 1 ) 2 = 2 x 3 − 2 x − 2 x 3 − 2 x ( x 2 − 1 ) 2 = − 4 x ( x 2 − 1 ) 2 \frac{d}{dx}\left[\frac{x^2 + 1}{x^2 - 1}\right] = \frac{(2x)(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}
d x d [ x 2 − 1 x 2 + 1 ] = ( x 2 − 1 ) 2 ( 2 x ) ( x 2 − 1 ) − ( x 2 + 1 ) ( 2 x ) = ( x 2 − 1 ) 2 2 x 3 − 2 x − 2 x 3 − 2 x = ( x 2 − 1 ) 2 − 4 x
Summary Schedule: Product & Quotient Rules
Rule
Formula
Use When...
Product Rule
d d x [ f ( x ) g ( x ) ] = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) d x d [ f ( x ) g ( x )] = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x )
Differentiating two functions multiplied together
Quotient Rule
d d x [ f ( x ) g ( x ) ] = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} d x d [ g ( x ) f ( x ) ] = [ g ( x ) ] 2 f ′ ( x ) g ( x ) − f ( x ) g ′ ( x )
Differentiating a fraction (division of functions)
Exercises
1. Constant multiple rule
Differentiate the following using constant multiple rule:
a) d d x [ 5 x 3 ] \frac{d}{dx}[5x^3] d x d [ 5 x 3 ]
b) d d x [ − 7 x ] \frac{d}{dx}[-7\sqrt{x}] d x d [ − 7 x ]
c) d d x [ 3 e x ] \frac{d}{dx}[3e^x] d x d [ 3 e x ]
2. Product rule
Differentiate the following using the product rule:
a) d d x [ x ⋅ ln ( x ) ] \frac{d}{dx}[x \cdot \ln(x)] d x d [ x ⋅ ln ( x )]
b) d d x [ e x ⋅ x 2 ] \frac{d}{dx}[e^x \cdot x^2] d x d [ e x ⋅ x 2 ]
c) d d x [ ( x 2 + 4 x ) ( x 3 − 1 ) ] \frac{d}{dx}[(x^2 + 4x)(x^3 - 1)] d x d [( x 2 + 4 x ) ( x 3 − 1 )]
3. Quotient rule
Differentiate the following using the quotient rule:
a) d d x [ 1 x ] \frac{d}{dx}\left[\frac{1}{x}\right] d x d [ x 1 ]
b) d d x [ x 2 + 1 x ] \frac{d}{dx}\left[\frac{x^2 + 1}{x}\right] d x d [ x x 2 + 1 ]
c) d d x [ ln ( x ) x 2 ] \frac{d}{dx}\left[\frac{\ln(x)}{x^2}\right] d x d [ x 2 l n ( x ) ]
4. Additional Practice
Differentiate:
a) d d x [ x 2 ⋅ ln ( 3 x ) ] \frac{d}{dx}[x^2 \cdot \ln(3x)] d x d [ x 2 ⋅ ln ( 3 x )]
b) d d x [ x 3 + 2 x x 2 + 1 ] \frac{d}{dx}\left[\frac{x^3 + 2x}{x^2 + 1}\right] d x d [ x 2 + 1 x 3 + 2 x ]
c) d d x [ e x x 3 + 2 ] \frac{d}{dx}\left[\frac{e^x}{x^3 + 2}\right] d x d [ x 3 + 2 e x ]
d) d d x [ ( 2 x 2 + 3 ) ( x 3 − 5 x ) ] \frac{d}{dx}[(2x^2 + 3)(x^3 - 5x)] d x d [( 2 x 2 + 3 ) ( x 3 − 5 x )]
e) d d x [ x 2 ( x − 1 ) x + 2 ] \frac{d}{dx}\left[\frac{x^2(x-1)}{x+2}\right] d x d [ x + 2 x 2 ( x − 1 ) ]
Hint: Expand numerator before applying quotient rule if easier.
5. Evaluate
Let f ( x ) = g ( x ) h ( x ) f(x) = \frac{g(x)}{h(x)} f ( x ) = h ( x ) g ( x ) . Suppose g ( 2 ) = 5 g(2) = 5 g ( 2 ) = 5 , g ′ ( 2 ) = 3 g'(2) = 3 g ′ ( 2 ) = 3 , h ( 2 ) = 1 h(2) = 1 h ( 2 ) = 1 , and h ′ ( 2 ) = − 4 h'(2) = -4 h ′ ( 2 ) = − 4 .
6. Additional practice
Follow and solve the exercises in this link .
Extra Advanced Exercises
1. Proof of Product and Quotient Rules
Using the limit definition of derivatives, show that:
(a) d d x ( f ( x ) g ( x ) ) = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) \frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x) d x d ( f ( x ) g ( x )) = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x )
(b) d d x ( f ( x ) g ( x ) ) = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} d x d ( g ( x ) f ( x ) ) = g ( x ) 2 f ′ ( x ) g ( x ) − f ( x ) g ′ ( x )
2. Water Flow in a Conical Container
Water is being poured into a conical tank such that the volume added per second is related to the height of the water x x x (in meters) and the radius of the water surface r r r by:
f ( x ) = x 2 ln ( x ) x + 1 f(x) = \frac{x^2 \ln(x)}{x + 1}
f ( x ) = x + 1 x 2 ln ( x )
Find how fast the volume is changing with respect to the height.
3. Nested Quotient of Products
Let
f ( x ) = ( x 2 + 1 ) ( x − 4 ) ( x + 2 ) ( x 2 − 1 ) f(x) = \frac{(x^2 + 1)(x - 4)}{(x + 2)(x^2 - 1)}
f ( x ) = ( x + 2 ) ( x 2 − 1 ) ( x 2 + 1 ) ( x − 4 )
(a) Differentiate f ( x ) f(x) f ( x )
(b) Simplify and identify where f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0
4. Pressure–Volume Constraint in a Gas
The gas pressure and volume are related by:
ln ( x y ) = x y \ln(xy) = \frac{x}{y}
ln ( x y ) = y x
Use implicit differentiation (and logarithmic identities) to find how volume y y y changes as pressure x x x increases.
5. Two Interrelated Functions
Let:
f ( x ) = g ( x ) h ( x ) , and g ( x ) h ( x ) = x 2 + 1 f(x) = \frac{g(x)}{h(x)}, \quad \text{and } g(x)h(x) = x^2 + 1
f ( x ) = h ( x ) g ( x ) , and g ( x ) h ( x ) = x 2 + 1
Use implicit differentiation on the product identity to find f ′ ( x ) f'(x) f ′ ( x ) .