Understanding the product rule for the derivative

When you first start to learn the concepts of differential calculus, you begin by learning how to take the derivative of the various functions. You learn that the derivative of sin (x) cos (x), the derivative of ax ^ n is anx ^ (n-1), and a number of other rules for the basic functions you saw all through Algebra and trigonometry. After learning about compensating for individual functions, see derivatives of products of these functions, which expands the range of functions that you can take the derivative of.

However, there is a big step up complexity when you move from taking the derivative of the basic functions of taking derivatives of products of functions. Because of this big step up in complexity is the process, many students feel grateful and really have a lot of problems understanding the material. Unfortunately, many instructors give students methods address these issues, but we do! Let's get started.

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

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

This type is what is called the rule of the product. This is more complex than any previous formulas for compensating you've seen up to this point, your calculus sequence. However, if you record each function you have to do with before attempting to write ƒ ' (x), then your speed and accuracy will improve significantly. So, step one is to write that a (x) what they are and what b (x). Then next to it, find the derivatives of a (x) and ' b ' (x). Once you have everything written, then there is nothing else to think about and you just fill in the blanks for the type of rule "product". This is all there is to it.

Let's use a tough example to show how easy it is for this procedure. Suppose we want to find the derivative of the following:

f (x) = (5sin (x) + 4 x-16) (3cos (x)-2 x + 4 x + 5)

Remember that step in determining what is a (x) and b (x). Clearly, a (x) = 5sin (x) + 4 x-16 x and b (x) = 3cos (x)-2 x + 4 x + 5, since these are the two functions that are multiplied together to form f (x). From our side of the Paper, only then we can write:

a (x) = 5sin (x) + 4 x-16 x
b (x) = 3cos (x)-2 x + 4 x + 5

By written by separate from each other, now we find the derivative of a (x) and b (x) individual just below it. Remember that these are basic functions, so that we know already how derivatives:

a ' (x) = 5cos (x) + 12-16
b ' (x) =-(x) 3sin-4 x + 4

With all this writes in an organized manner, you don't have to remember anything anymore! All work on this problem has ended. You need only to write these four functions in the correct sequence, which is given to us by the product rule.

Finally, you write that the basic form of the rule of product, f ' (x) = a (x) * b (x) + a (x) * b ' (x), and write the corresponding functions in place of a (x), a ' (x), b (x) and b (x). Returning above where we work: our problem

f ' (x) = a (x) * b (x) + a (x) * b ' (x)
f ' (x) = (5cos (x) + 12-16) * (3cos (x)-2 x + 4 x + 5) + (5sin (x) + 4 x-16 x) * ((x)-3sin-4 x + 4)

This is a very large derivative function, but when we organise our thinking efficiently, quickly and accurately to take compensating products regardless of how long is the initial function!