Finding Derivatives Made Easy! Product Rule of Differentiation
The product rule for differentiation calculates the derivative of two multiplied functions, combining each function's derivative with the other function in a sum. The product rule is crucial in calculus for handling complex functions, enabling efficient differentiation of products, and fundamental in physics and engineering applications.
Finding the derivative of product of two functions:
As mentioned earlier, we can’t find the derivative of the product of two function by multiplying the derivative of each individual function. This only works when dealing with addition/subtraction. Instead, we need to use the formula:
\( (f(x) \times g(x))’ = f'(x) \cdot g(x) + g'(x) \cdot f(x) \)
So basically, finding the derivative of first function and multiplying it by the second, and adding it to the derivative of second function multiplied by the first function.
You can sometimes avoid using product rule:
Sometimes it can be easier to simplify the given functions by multiplying them first, and taking the derivative of the resulting expression, meaning to avoid using product rule. Both methods yield the same answer, yet sometimes one method can be easier than the other.
For instance, the derivative of \( 2x(x^2) \)
First, using the product rule:
\( f(x) = 2x \cdot x^2 \Rightarrow f'(x) = 2x \cdot 2x + 2 \cdot x^2 = 4x^2 + 2x^2 = 6x^2 \)
Now, if simplifying first:
\( f(x) = 2x \cdot x^2 = 2x^3 \Rightarrow f'(x) = 3 \cdot 2x^{2} = 6x^2 \)
Product rule involving other functions:
Multiplication of other functions, like radicals and trigonometric functions, follows the same principle. For example:
\( f(x) = x \sin x \Rightarrow f'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x \)
Here is another example:
\( f(x) = \cos x \cdot x^{1/3} \Rightarrow f'(x) = -\sin x \cdot x^{1/3} + \frac{1}{3} \cos x \cdot x^{-2/3} \)
Finding the derivative of product of three functions:
Here is a more complex example, involving chain rule:
\( f(x) = \cos 2x \cdot (3x)^{1/3} \Rightarrow f'(x) = -2 \sin 2x \cdot (3x)^{1/3} + \cos 2x \cdot \frac{1}{3} \cdot (3x)^{-2/3} \cdot 3 \)
If you have three functions and want to apply the product rule for differentiation, the process involves a bit more complexity than with two functions.
- Differentiate the first function and multiply it by the unchanged second and third functions.
- Keep the first function unchanged, differentiate the second function, and multiply by the unchanged third function.
- Keep the first and second functions unchanged and differentiate the third function.
For the functions \( f(x), g(x), h(x) \) , the mathematical formula for the derivative of product of these functions, is:
\( f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \)
Here’s an example:
\( f(x) = x^2, \ g(x) = x, \ h(x) = x \)
\( f'(x) = 2x, \ g'(x) = 1, \ h'(x) = 1 \)
\(\Rightarrow f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) = 2x \cdot x \cdot x + x^2 \cdot 1 \cdot x + x^2 \cdot x \cdot 1 \)
\(= 2x^3 + x^3 + x^3 \)
\(= 4x^3 \)
Related to This Article
More math articles
- How to Find Vector Components?
- Top 10 Tips to Overcome CLEP College Algebra Anxiety
- Integral Approach to Cumulative Growth
- FREE AFOQT Math Practice Test
- 5th Grade NDSA Math Worksheets: FREE & Printable
- What Skills Do I Need for the CHSPE Math Test?
- How to Understand Random Sampling and Variation in Samples?
- 7th Grade PSSA Math Practice Test Questions
- FREE 7th Grade SBAC Math Practice Test
- How to Find Complex Roots of the Quadratic Equation?
What people say about "Finding Derivatives Made Easy! Product Rule of Differentiation - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.