In the second part of the formula, the upper case Greek letter Sigma ( Σ) is used to indicate that the derivative of the product of these functions is equal to the sum of something. The upper case Greek letter Pi ( Π) is then used to indicate that it is the derivative of a product - in this case, the product of the functions ƒ1( x) through ƒ n( x). The first part of the formula starts off by telling us that we are looking at the derivative of something, using the notation we are familiar with. It's just a formal way of saying how, for a collection of n functions, we can find the derivative of the product of those functions. ƒ n, we can use the following generalisation:ĭon't worry if this looks a bit frightening. For example, the product of the three functions u, v and w can be found using the following formula:įor the product of n functions ƒ1. We can use a general form of the product rule to find the derivative of the product of more than two functions. Now we find the differential for each of these functions: Supposing we want to find the derivative of y = ( x 2 - 4 x)(3 - 2 x 3). This is the result we obtained previously, as we would expect. Let's test the formula by applying it to the function ƒ( x) = x( x 2 + 1), for which we have already obtained the derivatives of the individual functions and their product: The derivative of the product of these two functions uv is given by: Fortunately, there is a simple formula (the product rule) that we can use to find the derivative of the product of two functions.Įssentially, the rule states that in order to find the derivative of the product of two functions, we take the first function multiplied by the derivative of the second function, and add it to the second function multiplied by the derivative of the first function. There will be occasions, however, when this will much more difficult or impossible to do. In this case, multiplying out the two functions was a relatively trivial exercise, and we were able to find the derivative of the function ƒ( x) = x( x 2 + 1) without difficulty. This is obviously very different from the result we got by simply multiplying together the derivatives of the functions ƒ(x) = x and ƒ( x) = x 2 + 1. Taking the derivative of x 3 + x, we get: Multiplying out the brackets in our original function, however, we get: Obviously, if we multiply these two derivatives together, the result will be x. We'll start by finding the derivative of each function separately: This function is the product of two functions, ƒ( x) = x and ƒ( x) = x 2 + 1. Suppose we need to find the derivative of the function ƒ( x) = x( x 2 + 1). An example should serve to illustrate the point. Unfortunately, it is not quite that simple. two functions multiplied together), we can simply multiply together the derivatives of each function. You might therefore be tempted to assume that, for a function that is the product of two functions (i.e. We know that we can find the differential of a polynomial function by adding together the differentials of the individual terms of the polynomial, each of which can be considered a function in its own right. The product rule gives us the derivative of the product of two (or more) functions.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |