### Video Transcript

Using the product rule, find the derivative of π₯ squared π to the power of negative π₯.

Well, we know that we can use the product rule because actually we have our function in the form π¦ equals π’π£. And what the product rule tells us is that if we have a function in this form, then the derivative is gonna be equal to π’ dπ£ dπ₯ plus π£ dπ’ dπ₯. So what that tells us is that π’, so the first part of our function, multiplied by the derivative of our π£ plus π£ multiplied by the derivative of our π’.

So the first thing we do with our question is actually identify what our π’ and π£ are gonna be. Our π’ is gonna be π₯ squared, our π£ is gonna be equal to π to the power of negative π₯. So therefore, our dπ’ dπ₯ is gonna be equal to two π₯. And we got that because we use the same rule that we do when weβre differentiating any term. And thatβs actually we multiply the coefficient, which was one, by the exponent, which was two, which gives us our two. And then, we reduce the exponent by one. So weβve got two minus one, which is one. So we get two π₯.

Okay, great, so now letβs find dπ£ dπ₯. Well, dπ£ dπ₯ is just gonna be equal to negative π to the power of negative π₯. So what we know about π is that if you actually differentiate π to the power of π₯, you get π to the power of π₯. Thatβs one of its sort of special properties.

But how we actually got our answer is using the chain rule cause we know that if you have π to the power of π of π₯, then thisβs gonna be equal to the derivative of π of π₯ multiplied by π to power of π of π₯. Well, in our case, our π of π₯ was negative π₯, where the derivative of negative π₯ is just negative one. So therefore, itβll be negative one multiplied by π to the power of negative π₯, which is what we got.

So now, what we can do is actually bring everything together using the product rule to actually find the derivative of π₯ squared π to the power of negative π₯. So what weβre first gonna have is π₯ squared multiplied by negative π to the power of negative π₯. And thatβs our π’ and our dπ£ dπ₯, so π’ multiplied by dπ£ dπ₯. And then, this is plus π to the power of negative π₯ multiplied by two π₯ because this is our π£ dπ’ dπ₯.

So then, if we actually tidied this up, we can actually say that the derivative of π₯ squared π to the power of negative π₯ is gonna be equal to two π₯ π to the power of negative π₯ minus π₯ squared π to the power of negative π₯.