# What Is A Derivative In Math latest 2023

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## Understanding the Product Rule for Derivatives

When you begin to learn the concepts of differential calculus, you begin by learning to take the derivatives of various functions. You learn that the derivative of sin(x) is cos(x), the derivative of ax^n is anx^(n-1), and a number of other rules for basis functions that you’ve seen throughout along algebra and trigonometry. After learning the derivatives of individual functions, you examine the derivatives of the products of those functions, which greatly expands the range of functions you can take the derivative of.

However, there is a big leap in complexity when you go from taking derivatives of basis functions to taking derivatives of products of functions. Due to this big leap in the complexity of the process, many students feel overwhelmed and have a lot of trouble really understanding the material. Unfortunately, many instructors don’t give students methods to solve these problems, but we do! Let’s start.

Suppose we have a function f(x) consisting of two regular functions multiplied together. Let’s call these two functions a(x) and b(x), which would mean that we have f(x) = a(x) * b(x). Now we want to find the derivative of f(x), which we call f'(x). The derivative of f(x) will look like this:

f'(x) = a'(x) * b(x) + a(x) * b'(x)

This formula is what we call the product rule. It is more complicated than all the previous formulas for derivatives that you will have seen so far in your calculation sequence. However, if you write every function you work with BEFORE you attempt to write f'(x), then your speed and accuracy will improve greatly. So the first step is to write down what a(x) is and what b(x) is. Then next to that find the derivatives a'(x) and b'(x). Once you’ve written it all down, there’s nothing more to think about and you just fill in the blanks for the product rule formula. That’s all we can say about it.

Let’s take a difficult example to show how easy this process is. Suppose we want to find the derivative of:

f(x) = (5sin(x) + 4x³ – 16x)(3cos(x) – 2x² + 4x + 5)

Remember that the first step is to identify what a(x) and b(x) are. Clearly, a(x) = 5sin(x) + 4x³ – 16x and b(x) = 3cos(x) – 2x² + 4x + 5, since these are the two functions multiplied together to form f(x). On the side of our paper, we simply write:

a(x) = 5sin(x) + 4x³ – 16x

b(x) = 3cos(x) – 2x² + 4x + 5

With that written separately from each other, we now find the derivative of a(x) and b(x) individually just below. Remember that these are basic functions, so we already know how to take their derivatives:

a'(x) = 5cos(x) + 12x² – 16

b'(x) = -3sin(x) – 4x + 4

With everything written in an organized way, we don’t have to remember anything! All work for this issue is complete. It is enough to write these four functions in the correct order, which is given to us by the rule of the product.

Finally, you write the basic format of the product rule, f'(x) = a'(x) * b(x) + a(x) * b'(x), and write the respective functions instead of a (x), a'(x), b(x) and b'(x). So back to where we are working our problem, we have:

f'(x) = a'(x) * b(x) + a(x) * b'(x)

f'(x) = (5cos(x) + 12x² – 16) * (3cos(x) – 2x² + 4x + 5) + (5sin(x) + 4x³ – 16x) * (-3sin(x) – 4x + 4)

It’s a very long derivative function, but when we organize our thinking efficiently, we can quickly and accurately take derivatives of products, no matter how long the original function is!

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