# 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.

1. Differentiate the first function and multiply it by the unchanged second and third functions.
2. Keep the first function unchanged, differentiate the second function, and multiply by the unchanged third function.
3. 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:

$$f(x) = x^2, \ g(x) = x, \ h(x) = x$$

$$f'(x) = 2x, \ g'(x) = 1, \ h'(x) = 1$$

$$\Rightarrow f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) = 2x \cdot x \cdot x + x^2 \cdot 1 \cdot x + x^2 \cdot x \cdot 1$$

$$= 2x^3 + x^3 + x^3$$

$$= 4x^3$$

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