Sure! Here are a few examples of how to use the product rule:
Example 1:
Find the derivative of f(x) = x^3 * sin(x).
Solution:
Let u = x^3 and v = sin(x).
Using the product rule: f'(x) = u'v + uv'
Next, find the derivatives of u and v:
u' = 3x^2 (the power rule for differentiation)
v' = cos(x) (the derivative of sin(x) is cos(x))
Now substitute these values into the product rule:
f'(x) = (3x^2)(sin(x)) + (x^3)(cos(x))
So the derivative of f(x) is: f'(x) = 3x^2sin(x) + x^3cos(x).
Example 2:
Find the derivative of f(x) = (2x^2 + 3)(4x - 5).
Solution:
Let u = 2x^2 + 3 and v = 4x - 5.
Using the product rule: f'(x) = u'v + uv'
Next, find the derivatives of u and v:
u' = 4x (the power rule for differentiation)
v' = 4 (the derivative of 4x is 4)
Now substitute these values into the product rule:
f'(x) = (4x)(4x - 5) + (2x^2 + 3)(4)
Simplifying, we get: f'(x) = 8x^2 - 20x + 8x^2 + 12
Combining like terms, we get: f'(x) = 16x^2 - 20x + 12.
These are just a couple of examples of how to use the product rule. Just remember to identify the two functions (u and v), find their derivatives (u' and v'), and apply the product rule formula: f'(x) = u'v + uv'.
product rule examples
3 answers
with just numbers
Certainly! Here are a couple of examples using numbers only:
Example 1:
Find the derivative of f(x) = 2x * 3.
Solution:
Let u = 2x and v = 3.
Using the product rule: f'(x) = u'v + uv'
Next, find the derivatives of u and v:
u' = 2 (the derivative of 2x is 2)
v' = 0 (the derivative of a constant is always zero)
Now substitute these values into the product rule:
f'(x) = (2)(3) + (2x)(0)
Simplifying, we get: f'(x) = 6 + 0
So the derivative of f(x) is: f'(x) = 6.
Example 2:
Find the derivative of f(x) = (4x - 3)(2x + 5).
Solution:
Let u = 4x - 3 and v = 2x + 5.
Using the product rule: f'(x) = u'v + uv'
Next, find the derivatives of u and v:
u' = 4 (the derivative of 4x is 4)
v' = 2 (the derivative of 2x is 2)
Now substitute these values into the product rule:
f'(x) = (4)(2x + 5) + (4x - 3)(2)
Simplifying, we get: f'(x) = 8x + 20 + 8x - 6
Combining like terms, we get: f'(x) = 16x + 14.
These examples show how to apply the product rule when dealing with simple numerical expressions.
Example 1:
Find the derivative of f(x) = 2x * 3.
Solution:
Let u = 2x and v = 3.
Using the product rule: f'(x) = u'v + uv'
Next, find the derivatives of u and v:
u' = 2 (the derivative of 2x is 2)
v' = 0 (the derivative of a constant is always zero)
Now substitute these values into the product rule:
f'(x) = (2)(3) + (2x)(0)
Simplifying, we get: f'(x) = 6 + 0
So the derivative of f(x) is: f'(x) = 6.
Example 2:
Find the derivative of f(x) = (4x - 3)(2x + 5).
Solution:
Let u = 4x - 3 and v = 2x + 5.
Using the product rule: f'(x) = u'v + uv'
Next, find the derivatives of u and v:
u' = 4 (the derivative of 4x is 4)
v' = 2 (the derivative of 2x is 2)
Now substitute these values into the product rule:
f'(x) = (4)(2x + 5) + (4x - 3)(2)
Simplifying, we get: f'(x) = 8x + 20 + 8x - 6
Combining like terms, we get: f'(x) = 16x + 14.
These examples show how to apply the product rule when dealing with simple numerical expressions.