Differentiation using multiple rules
Product rule to find derivative of product of three functions
What I want to do in this video is think about how we can take the derivative of an expression that can be viewed as a product not of two functions but of three functions. And we're going to do it using what we know of the product rule. And the way we could think about it is we can view this as the product, first, of two functions, of this function here and then that function over there. And then separately take the derivative. So if we just view the standard product rule, it tells us that the derivative of this thing will be equal to the derivative of f of x-- let me close it with a white bracket-- times the rest of the function. So times g of x-- let me close it with the-- times g of x times h of x times plus just f of x times the derivative of this thing. Times the derivative with respect to x of g of x times h of x. Let me write that a little bit neater. But what is this thing right over here going to be equal? Well we can apply the product rule again. So here, I'm just focusing on this part right over here. The derivative of this is just going to be g prime of x times h of x plus g of x times the derivative of h, times h prime of x. So everything that we had the derivative of g of x times h of x is this stuff right over here. And so we're going to multiply that times f of x. So let's rewrite all of this stuff here. So this first term right over here we can rewrite. So all of this is going to be equal to f prime of x-- that's that right over there-- times g of x times h of x plus-- And now we're going to distribute this f of x. So it's f of x times this plus f of x times this. So f of x times this is f of x times g prime of x, the derivative of g, g prime of x, times h of x. Let me do that in that white color. And then finally, f of x times this is just f of x times g of x times h prime of x. And this was a pretty neat result. Essentially, we can view this as the product rule where we have three, where we could have our expression viewed as a product of three functions. Now we have three terms. In each of these terms, we take a derivative of one of the functions and not the other two. Here we took the derivative of f. Here we took the derivative of g. Here we took the derivative of h. And you can imagine, if you had the product of four functions here, you would have four terms. In each of them you'd be taking a derivative of one of the functions. If you had n functions here, then you would have n terms here. And in each of them you would take the derivative of one of the functions. So this is kind of a neat result.