Product and Quotient Rules
Product Rule
Calculus has two simple formulas to find the derivatives of the sums and differences of functions. The derivative of a product of
functions is not the same thing as taking the derivative of each term and then multiplying them together.
The Product Rule states: If f(x) = u(x) . v(x), where u and v are differentiable functions of x, then f ' (x) = u(x) . v'(x) + v(x) . u'(x)
The preceding formula says that the derivative of a product of two functions is the first term times the derivative of the second term plus the second term
times the derivative of the first term.
Example: Use the Product Rule to find the derivative of f (x) = (x 2 + x)(2x +1)
Identify u(x), u'(x), v(x), and v'(x)
Step 1
u(x) = x 2 +x , v(x) = 2x +1
u'(x) = 2x +1 , v'(x) = 2
Step 2
Plug in the values identified in Step 1 into the formula
f '(x) = u(x) . v'(x) + v(x) . u'(x)
f '(x) = (x 2 + x)(2) + (2x + 1)(2x +1)
f '(x) = 2 x 2 + 2x + 4 x 2 + 2x + 2x +1
f '(x) = 6 x 2 + 6x + 1
Product Rule → → →
Substitute the values found in Step 1 → → →
Multiply the terms out → → →
Simplify
→→→
Therefore, using the Product rule, the derivative of the function (x 2 + x)(2x +1) is 6 x 2 + 6x + 1.
Example 2:
Use the Product Rule to find the derivative of f(x) = xsinx
Step 1
Identify u(x), u’(x), v(x), v’(x)
The derivative of x is 1 and the derivative of sinx is cosx
u(x) = x, v(x) = sinx
u ’(x) = 1, v ’(x) = cosx
Step 2
Plug in the values identified in Step 1 into the formula
f ' (x) = u(x) . v'(x) + v(x) . u'(x)
f ’(x) = (x)(cosx) + (sinx)(1)
f ’(x) = xcosx + sinx
Therefore, using the Product rule, the derivative of the function (xsinx) is (xcosx + sinx)
Product Rule → → →
Substitute the values found in Step 1 → → →
Multiply the terms out → → →
Example 3:
Use the Product Rule to find the derivative of f(x) = xex
Step 1
Identify u(x), u ’(x), v(x), v ’(x)
The derivative of x is 1 and the derivative of ex is ex
u(x) = x, v(x) = ex
Step 2
u ’(x) = 1, v ’(x) = ex
Plug in the values identified in Step 1 into the formula
Substitute the values found in Step 1 → → →
f ' (x) = u(x) . v'(x) + v(x) . u'(x)
f ’(x) = x ex + ex (1)
Multiply the terms out → → →
f ’(x) = x ex + ex
Product Rule → → →
The derivative of the function (xex ) is (x ex + ex).
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Quotient Rule.
The quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives
exist.
u ( x)
The Quotient Rule states: If f(x) =
, where u and v are differentiable functions of x, and v(x) ≠ 0, then f'(x) =
v( x)
v( x) ⋅ u' ( x) − u( x) ⋅ v' ( x)
[v( x)]2
The preceding formula says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the
derivative of the denominator, all divided by the square of the denominator.
Example: Use the quotient rule to find f '(x) if f(x) =
Step 1
Identify u(x), u '(x), v(x), v '(x)
x 2 − 4x
x+5
u(x) = x 2 - 4x, v(x) = x + 5
u'(x) = 2x - 4, v '(x) = 1
Step 2
Plug in the values identified in Step 1 into the formula
Quotient Rule
→→→
f '(x) =
f '(x) =
Substitute the values found in step 1 →→→
f '(x) =
Multiply the terms out →→→
[v( x)]2
( x + 5)(2 x − 4) − ( x 2 − 4 x)(1)
( x + 5) 2
2 x 2 − 4 x + 10 x − 20 − x 2 + 4 x
( x + 5) 2
f '(x) =
Simplify →→→
v( x) ⋅ u ' ( x) − u ( x) ⋅ v' ( x)
x 2 + 10 x − 20
( x + 5) 2
Therefore, using the Quotient rule, the derivative of the function
2
x 2 − 4x
is x + 10 x −2 20 .
x+5
( x + 5)
Example 2: Use the quotient rule to find the derivative of f(x) =
4x 2 − 3
x 2 +1
Step 1
Identify u(x), u '(x), v(x), v '(x)
u(x) = 4 x 2 − 3 , v(x) = x 2 + 1
u’(x) = 8 x ,
v ‘(x) = 2x
Step 2
Plug in the values identified in Step 1 into the formula
Quotient Rule
→→→
f '(x) =
Substitute the values found in step 1 →→→
f ‘(x) =
v( x) ⋅ u ' ( x) − u ( x) ⋅ v' ( x)
[v( x)]2
( x 2 + 1)(8 x) − (4 x 2 − 3)(2 x)
( x 2 + 1) 2
8x 3 + 8x − 8x 3 + 6 x
f ‘(x) =
( x 2 + 1) 2
Multiply the terms out →→→
f ‘(x) =
Simplify →→→
Therefore, using the Quotient rule, the derivative of the function
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4x
( x + 1) 2
2
4x 2 − 3
4x
is 2
.
2
( x + 1) 2
x +1
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