# Video: Differentiating a Combination of Root and Polynomial Functions Using the Product Rule

Find the first derivative of the function 𝑓(𝑥) = (2𝑥⁴ + 𝑥 − 5)(𝑥² + 3√𝑥 − (3/𝑥)).

07:26

### Video Transcript

Find the first derivative of the function 𝑓 of 𝑥 is equal to two 𝑥 to the power of four plus 𝑥 minus five multiplied by 𝑥 squared plus three root 𝑥 minus three over 𝑥.

So the first thing we’re actually gonna do is actually rewrite our function. But what we’re gonna do is actually remove the terms that aren’t exponents and actually change them for exponents of 𝑥. So therefore when we do that, we’re gonna get that 𝑓 of 𝑥 is equal to two 𝑥 to the power of four plus 𝑥 minus five multiplied by 𝑥 squared plus three 𝑥 to the power of a half — and that’s because root 𝑥 is equal to 𝑥 to the power of a half — minus three 𝑥 to the power of negative one, and that’s because one over 𝑥 is equal to 𝑥 to the power of negative one.

Okay great! But how are we actually gonna differentiate this. Well we’re actually gonna use the product rule. And we can use the product rule because our function is in the form 𝑦 equals 𝑢𝑣, because actually we have one thing multiplied by another. And what the product rule tells us is that actually there’s a way of finding the first derivative. And that is that if you can find the first derivative, so d𝑦 d𝑥, by having 𝑢 multiplied by d𝑣 d𝑥 plus 𝑣 multiplied by d𝑢 d𝑥.

So what this means is actually our 𝑢 multiplied by the derivative of our 𝑣 plus our 𝑣 multiplied by the derivative of our 𝑢. So let’s now apply this and use it to actually find the first derivative of our function. So the first thing we’ve done is actually identified what our 𝑢 and our 𝑣 are going to be. Our 𝑢 is gonna be two 𝑥 to the power of four plus 𝑥 minus five, and our 𝑣 is going to be 𝑥 squared plus three 𝑥 to the power of a half minus three 𝑥 to the power of negative one.

So now what we’re gonna do is actually differentiate both 𝑢 and 𝑣. So d𝑢 d𝑥 is gonna be equal to, so we get eight 𝑥 cubed plus one for the derivative of 𝑢. And we get that using the normal differentiation rule. So for instance, we’ll think about the first term. We got that because actually we’ve got two multiplied by four, so our coefficient multiplied by our exponent, which gives us our eight. And then we reduce the exponent by one, which gives us our 𝑥 cubed. Okay great! Now let’s move on to d𝑣 d𝑥. Well for d𝑣 d𝑥, we’re actually gonna get two 𝑥 plus three over two 𝑥 to the power of negative half plus three 𝑥 to the power of negative two.

So we actually got that using the normal set of rules for differentiation. It’s worth pointing out this part here. And we’ve actually got three over two. And we got three over two because this is three multiplied by a half, which gives us three over two or one and half. And we’ve got 𝑥 to the power of negative a half. And we actually got the negative a half because if you’ve got a half minus one, that gives us negative a half. Be careful again with negatives, okay? So what we can do now is actually apply our product rule to actually find the first derivative of our function.

So we’re gonna have the first derivative is equal to, well first of all, two 𝑥 to the power of four plus 𝑥 minus five multiplied by two 𝑥 plus three over two 𝑥 to the power of negative a half plus three 𝑥 to the power of negative two. And the reason we get that is because actually what we’ve got is our 𝑢 and our d𝑣 d𝑥. And then this is plus we’ve got 𝑥 squared plus three 𝑥 to the power of a half minus three 𝑥 to the power of negative one 𝑥 multiplied by eight 𝑥 cubed plus one. And this is because this is our d𝑣 d𝑢 d𝑥.

So now the next stage is to actually expand our parentheses. So we’re gonna get four 𝑥 to the power of five because we’ve got two 𝑥 to the power of four multiplied by two 𝑥. And then plus three 𝑥 to the power of seven over two, and that’s because we have two multiplied by three over two. So that gives us three, so two one and a half, so it’s three. And then we’ve got 𝑥 to the power of four multiplied by 𝑥 to the power of negative a half. Well if we actually add the exponents. We’re gonna have four, which is the same as eight over two, minus one over two, which gives us seven over two.

Then plus six 𝑥 squared, then plus two 𝑥 squared, then plus three over two 𝑥 to the power of a half. And we got that because you got 𝑥, so that’s like 𝑥 to the power of one, then you multiply 𝑥 to the power one by 𝑥 to the power of negative a half. Well one minus a half gives us a half. And then finally for this one, plus three 𝑥 to the power of negative one. Okay, great! That’s the first two terms multiplied. Let’s move on to the last term in our first parentheses. So we get minus 10𝑥 minus 15 over two 𝑥 to the power of negative a half and then minus 15𝑥 to the power of negative two.

Okay, great! That’s the first pair of parentheses multiplied. And then this is add eight 𝑥 to the power of five plus 𝑥 squared. So then this gives us 24𝑥 to the power of seven over two plus three 𝑥 to the power of a half. And that’s because we actually got the 24𝑥 to the power of seven over two, because we get three multiplied by eight which gives us 24 and then 𝑥 to the power of a half multiplied by 𝑥 to the power of three. So you add the exponents, so you get three and a half, which same as seven over two.

And then finally, we multiply the last term. In each of the parentheses, we get negative 24𝑥 squared minus three 𝑥 to the power of negative one. As you can see, there are a lot of terms. This is quite a bit of a long expansion. So be very careful when you multiply each term just to make sure that actually each term is multiplied correctly. Okay, so now what we need to do is actually simplify. And we’re actually gonna do that in steps.

So first of all, we’re gonna look to the 𝑥 to the power of five terms. Well we have four plus eight, which gives us 12𝑥 to the power of five. So we’re gonna work in descending orders of 𝑥. So therefore, the next one is going to be 𝑥 to the power of seven over two. So we get plus 27𝑥 to the power of seven over two. We now move on to the 𝑥 squared terms. And we’ve got positive six 𝑥 squared plus two 𝑥 squared, which gives us eight 𝑥 squared. Then we’ve got add another 𝑥 squared, which gives us nine 𝑥 squared. Then we subtract 24𝑥 squared, which gives us negative 15𝑥 squared.

And then our next term is plus nine over two 𝑥 to the power of a half — and that’s because we had three over two, so one and a half 𝑥 to the power of a half plus three 𝑥 to the power of a half, which gives us four and a half 𝑥 to the power of a half, which is nine over two 𝑥 to the power of a half — minus 10𝑥. And then we move in to the negative powers. So we’ve got minus 15 over two 𝑥 to the power of negative a half. Well then if we have a look, we actually have 𝑥 to the power of negative one terms. However, we’ve got positive three 𝑥 to the power of negative one, minus three 𝑥 to the power of negative one, so therefore these cancel each other out. So therefore finally what we have is minus 15𝑥 to the power of negative two.

Okay, great! So we’ve now actually simplified what we’re gonna do now is actually put it back into its original form. So we’re going to remove some of the exponents where we can actually have roots. So therefore, we can say that the first derivative of the function 𝑓 of 𝑥 is equal to two 𝑥 to the power four plus 𝑥 minus five multiplied by 𝑥 squared plus three root 𝑥 minus three over 𝑥 is 12𝑥 to the power of five plus 27𝑥 to the power of three multiplied by root 𝑥 minus 15𝑥 squared plus nine over two root 𝑥 minus 10𝑥 plus 15 over two root 𝑥 minus 15 over 𝑥 squared.