Find A Formula For F 1 X If F X Understanding the Product Rule for Derivatives

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Understanding the Product Rule for Derivatives

When you first start learning the concepts of differential calculus, you start learning how to take the derivatives of various functions. You learned that the derivative of sin(x) is cos(x), that the derivative of ax^n is anx^(n-1), and many other rules for basic operations that you’ve seen through algebra and trigonometry. After learning about derivatives for individual functions, you look at the derivatives of products of these functions, which greatly expands the range of functions that you can derive.

However, there is a big step up in complexity when you go from taking derivatives of basic functions to taking derivatives of products of functions. Because of this huge step in how complicated the process is, many students feel overwhelmed and have a lot of trouble really understanding the material. Unfortunately, many instructors don’t give students ways to address these issues, but we do! let’s begin

Suppose we have a function f(x) that consists of two regular functions multiplied together. Let’s call these two functions a(x) and b(x), which means 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 production law. This is more complicated than any of the previous formulas for derivatives that you’ve seen up to this point in your calculus sequence. However, if you write every function you’re dealing with before you try to write f'(x), your speed and accuracy will be greatly improved. 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 them all down, there’s nothing more to think about, and you fill in the blanks for the product rule formula. It’s all there.

Let’s use 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 those two functions are 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

Since they are written separately from each other, we now find the derivatives of a(x) and b(x) individually below them. 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

Everything is written organically, we don’t have to remember anything anymore! All work is done for this problem. We must write these four functions in the correct order, which is given to us by the product rule.

Finally, you write the basic form of the product rule, f'(x) = a'(x) * b(x) + a(x) * b'(x), and substitute the corresponding functions for a. (x), a'(x), b(x), and b'(x). So back to where we’re working, here’s our problem:

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)

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

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