# Question Video: Differentiating Natural Exponential Functions Using the Chain Rule Mathematics • Higher Education

Determine the derivative of π(π₯) = β5π^(β3π₯Β² + 3π₯).

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### 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, weβre first gonna rewrite it. And I have actually written it as the derivative of our function is equal to negative five multiplied by π ππ₯ of π to the power of negative three π₯ squared plus three π₯. The reason Iβve 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 ππ¦ ππ₯ or our derivative is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. 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 weβre gonna have as our π¦ and our π’. Well, first of all, π’ is actually going to be our exponent. So itβs 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, weβve got π¦ in a function of π’ and π’ is a function of π₯.

So now what I need to do is actually find ππ’ ππ₯ and ππ¦ ππ₯. Weβre gonna start with ππ’ ππ₯. So weβre 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, Iβm just gonna pick the first term. And we had our exponent two multiplied by coefficient negative three. So it gives us negative six. And weβve 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 letβs 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₯, weβre just going to get π to the power of π₯.

Okay, great! So weβve now done that, letβs use the chain rule and put it all together. So therefore, we can say that ππ¦ ππ₯ 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 thatβs ππ¦ ππ’, and then multiplied by negative six π₯ plus three, because that was our ππ’ ππ₯.

Now letβs tidy this up a bit. So when we actually tidy up, we get 30π₯. And thatβs 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 havenβt 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₯.