### Video Transcript

Determine the derivative of π π₯ equals negative five π to the power of negative three π₯ squared plus three π₯.

Now if you want to actually differentiate this function, weβre first gonna rewrite it. And I have actually written it as the derivative of our function is equal to negative five multiplied by π ππ₯ of π to the power of negative three π₯ squared plus three π₯. The reason Iβve done this is because, actually, the negative five is just a common factor. So we can actually deal with that separately.

So now what we need to do is actually differentiate π to the power of negative three π₯ squared plus three π₯. Now in order to actually differentiate this, what we can use is the chain rule. So what the chain rule actually states is that ππ¦ ππ₯ or our derivative is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. And we can use this when π¦ is in a function of π’ and π’ is in a function of π₯.

So the first thing we need to do is ask ourselves what are π¦ and π’. Well, we need to decide what weβre gonna have as our π¦ and our π’. Well, first of all, π’ is actually going to be our exponent. So itβs gonna be negative three π₯ squared plus three π₯. And therefore, π¦ is gonna be equal to π to the power of π’. So as we said was required to actually use the chain rule, weβve got π¦ in a function of π’ and π’ is a function of π₯.

So now what I need to do is actually find ππ’ ππ₯ and ππ¦ ππ₯. Weβre gonna start with ππ’ ππ₯. So weβre gonna differentiate negative three π₯ squared plus three π₯. And when we do that, we get negative six π₯ plus three. As a quick reminder of how we actually differentiate that, Iβm just gonna pick the first term. And we had our exponent two multiplied by coefficient negative three. So it gives us negative six. And weβve got π₯ to the power of two minus one cause we subtract one from the exponent, which just gives us π₯ to the power of one, or just π₯. So we get negative six π₯.

Okay, great! We differentiated that. Now letβs move on and differentiate π¦ equals π to the power of π’. Well, when we differentiate π to the power of π’, we actually just get π to the power of π’. And we get that because π is a special number for which the gradient of π to the power of π₯ is just π to the power of π₯. And therefore, if the gradient of π to the power of π₯ is π to the power of π₯, and that means when we differentiate π to the power of π₯, weβre just going to get π to the power of π₯.

Okay, great! So weβve now done that, letβs use the chain rule and put it all together. So therefore, we can say that ππ¦ ππ₯ is equal to negative five, because that was our constant that we actually took out in the beginning, multiplied by π to the power of π’, because thatβs ππ¦ ππ’, and then multiplied by negative six π₯ plus three, because that was our ππ’ ππ₯.

Now letβs tidy this up a bit. So when we actually tidy up, we get 30π₯. And thatβs because we had negative five multiplied by negative six π₯ and then minus 15, because we had negative five multiplied by three. Then this is multiplied by π to the power of π’.

But have we finished? Well, no, we havenβt actually finished because our differential is still in terms of π₯ and π’. And we actually just want it in terms of π₯. So what we need to do now is substitute back in our value for π’. Well then, therefore, we can say that the derivative of negative five π to the power of negative three π₯ squared plus three π₯ is equal to 30π₯ minus 15 multiplied by π to the power of negative three π₯ squared plus three π₯.