Video: Differentiating Natural Exponential Functions Using the Chain Rule

Determine the derivative of 𝑔(π‘₯) = βˆ’5𝑒^(βˆ’3π‘₯Β² + 3π‘₯).

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Video Transcript

Determine the derivative of 𝑔 π‘₯ equals negative five 𝑒 to the power of negative three π‘₯ squared plus three π‘₯.

Now if you want to actually differentiate this function, we’re first gonna rewrite it. And I have actually written it as the derivative of our function is equal to negative five multiplied by 𝑑 𝑑π‘₯ of 𝑒 to the power of negative three π‘₯ squared plus three π‘₯. The reason I’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 𝑒 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 𝑑𝑦 𝑑π‘₯ or our derivative is equal to 𝑑𝑦 𝑑𝑒 multiplied by 𝑑𝑒 𝑑π‘₯. And we can use this when 𝑦 is in a function of 𝑒 and 𝑒 is in a function of π‘₯.

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

So now what I need to do is actually find 𝑑𝑒 𝑑π‘₯ and 𝑑𝑦 𝑑π‘₯. We’re gonna start with 𝑑𝑒 𝑑π‘₯. So we’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, I’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 we’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 let’s move on and differentiate 𝑦 equals 𝑒 to the power of 𝑒. Well, when we differentiate 𝑒 to the power of 𝑒, we actually just get 𝑒 to the power of 𝑒. And we get that because 𝑒 is a special number for which the gradient of 𝑒 to the power of π‘₯ is just 𝑒 to the power of π‘₯. And therefore, if the gradient of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯, and that means when we differentiate 𝑒 to the power of π‘₯, we’re just going to get 𝑒 to the power of π‘₯.

Okay, great! So we’ve now done that, let’s use the chain rule and put it all together. So therefore, we can say that 𝑑𝑦 𝑑π‘₯ is equal to negative five, because that was our constant that we actually took out in the beginning, multiplied by 𝑒 to the power of 𝑒, because that’s 𝑑𝑦 𝑑𝑒, and then multiplied by negative six π‘₯ plus three, because that was our 𝑑𝑒 𝑑π‘₯.

Now let’s tidy this up a bit. So when we actually tidy up, we get 30π‘₯. And that’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 𝑒 to the power of 𝑒.

But have we finished? Well, no, we haven’t actually finished because our differential is still in terms of π‘₯ and 𝑒. And we actually just want it in terms of π‘₯. So what we need to do now is substitute back in our value for 𝑒. Well then, therefore, we can say that the derivative of negative five 𝑒 to the power of negative three π‘₯ squared plus three π‘₯ is equal to 30π‘₯ minus 15 multiplied by 𝑒 to the power of negative three π‘₯ squared plus three π‘₯.

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