# Video: Differentiating a Combination of Functions Involving Exponential Functions Using the Product Rule

Determine the derivative of 𝑦 = 𝑒^(−5𝑥) 𝑥².

04:19

### Video Transcript

Determine the derivative of 𝑦 equals 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 squared.

Now to enable us to actually determine the derivative, what we’re gonna use is the product rule. And we can actually use the product rule when we have our function in the form 𝑦 equals 𝑢𝑣. And if we take a look at our function, it’s actually in this form. So how does the actual product rule work? Well the product rule tells us that the derivative is equal to 𝑢 𝑑𝑣 𝑑𝑥 plus 𝑣 𝑑𝑢 𝑑𝑥.

So what this means is 𝑢 multiplied by the derivative of 𝑣 plus 𝑣 multiplied by the derivative of 𝑢. Okay, great! So now that we know the product rule, let’s use it to actually determine our derivative. So in order to actually determine our derivative, first of all what we need to do is to decide what 𝑢 and 𝑣 are. So 𝑢 is gonna be equal to 𝑒 to the power of negative five 𝑥, and 𝑣 is gonna be equal to 𝑥 squared.

Next, what we want to do is actually differentiate our 𝑢 and 𝑣. So I’m gonna start with 𝑣 because this is more straightforward. So we can say that d𝑣 d𝑥 is going to be equal to the derivative of 𝑥 squared. Well this is gonna be equal to two 𝑥, just remind us how we did that. So our exponent multiplied by our coefficient, so two multiplied by one, and then it’s 𝑥 to the power of, and then we reduce our exponent by one, so two minus one, which just be one. So we get two 𝑥.

So now we can move on to 𝑢. So if we wanna find d𝑢 d𝑥, well we’re actually gonna have to use is a general rule to help us here as well. And that’s because 𝑢 is in the form 𝑦 is equal to 𝑒 to the power of 𝑓 of 𝑥. And our rule tells us, if we have it in this form, then what we have is that the derivative is going to be equal to the derivative of 𝑓 of 𝑥 multiplied by 𝑒 to the power of 𝑓 of 𝑥. And this actually comes from an adaptation of the chain rule.

Okay, great! So we can use this to actually find out what d𝑢 d𝑥 is going to be. Well first of all, it is gonna be negative five. And that’s because if you differentiate negative five 𝑥, you get negative five. And then this is gonna be multiplied by 𝑒 to the power of negative five 𝑥. So great we now have d𝑢 d𝑥 and d𝑣 d𝑥. So now what we can do is actually go back to our product rule to actually find the derivative of our function.

So first of all, we’re gonna have 𝑒 to the power of negative five 𝑥 because that’s our 𝑢. And then this is gonna be multiplied by two 𝑥. And that’s because this is our 𝑑𝑣 𝑑𝑥. And then this is gonna be plus 𝑥 squared, which is our 𝑣, and then multiplied by negative five 𝑒 to the power of negative five 𝑥 because this is our 𝑑𝑢 𝑑𝑥. Okay, great! So now let’s rearrange this. And when we do that, we get two 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 minus five 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 squared.

Okay, so now what we can do is actually simplify this by taking out factors. So when we do that, there’s actually gonna be two results that we can have. So I’m gonna give you both of those. So the first one that we can find is if we take out 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 as a factor of each of our terms. So if we do that, then inside the parentheses we’re gonna get two minus five 𝑥. And this is actually our derivative simplified fully. So therefore, we can actually say that the derivative of 𝑦 equals 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 squared is 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 multiplied by two minus five 𝑥.

Okay, as I said, there’s actually another way that we can actually leave our answer. And we get this if we take out negative 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 as a factor this time instead. And we do that so that we can actually have the 𝑥 term as the first term in our parentheses this time. So what we get is the derivative is actually equal to negative 𝑒 to the power of negative five 𝑥 multiplied by 𝑥, then multiplied by five 𝑥 minus two. Okay, great! So we’ve actually determined the derivative of our function. And we did that using the product rule and then an adaptation of the chain rule.