Video: Differentiating Exponential Functions Using the Chain Rule

Video Transcript

Find 𝑑𝑦 𝑑𝑥 if 𝑦 is equal to four to the power of nine 𝑥 squared minus four 𝑥 plus eight.

Now, in order to actually find 𝑑𝑦 𝑑𝑥, what we’re gonna do is actually we’re gonna use the chain rule. And the chain rule states that 𝑑𝑦 𝑑𝑥 is equal to 𝑑𝑦 𝑑𝑢 multiplied by 𝑑𝑢 𝑑𝑥. And this is when 𝑦 is equal to a function of 𝑢 and 𝑢 is equal to a function of 𝑥. Okay, so we now got the chain rule. Let’s apply it to our problem to actually solve and find 𝑑𝑦 𝑑𝑥.

So first of all, we need to identify our 𝑢. So that’s gonna be nine 𝑥 squared minus four 𝑥 plus eight. And therefore, our 𝑦 is gonna be equal to four to the power of 𝑢. Okay, that’s the first step. Now, what we need to do is actually differentiate to find 𝑑𝑦 𝑑𝑢 and 𝑑𝑢 𝑑𝑥.

I’m gonna start by finding 𝑑𝑢 𝑑𝑥 and I will do that by differentiating the expression nine 𝑥 squared minus four 𝑥 plus eight. And this is just gonna give us 18𝑥 minus four. We’ve done that the normal way. So we just differentiated or highlighted that with the first term. So what we did is we multiplied the exponent by the coefficient. So two multiplied by nine is 18 and then we reduced the exponent by one — so from two to one. So we just get 18𝑥.

So great, we’ve differentiated that. Now, let’s move on and find 𝑑𝑦 𝑑𝑢. So now to find 𝑑𝑦 𝑑𝑢, what we’re gonna have to do is actually differentiate four to the power of 𝑢. And in order to do this, we’re actually gonna use the general rule which is that if we’re gonna differentiate 𝑎 to the power of 𝑥, then we’re gonna get an answer of 𝑎 to the power of 𝑥 multiplied by the natural logarithm of 𝑎. So therefore, as our result, we’re gonna get four to power of 𝑢 because that’s our 𝑎 and then multiply it by the natural logarithm of four because again that’s our 𝑎 because it’s 𝑎 to the power of 𝑥 ln 𝑎.

Great, so we’ve now differentiated both parts. We can use the chain rule to put them together to find 𝑑𝑦 𝑑𝑥. So now, we’re actually gonna apply the chain rule, which tells us that 𝑑𝑦 𝑑𝑥 is equal to 𝑑𝑦 𝑑𝑢 multiplied by 𝑑𝑢 𝑑𝑥. So we’re gonna get 𝑑𝑦 𝑑𝑥 is equal to four to the power of 𝑢 ln four and that’s because that was our 𝑑𝑦 𝑑𝑢 and then multiplied by 18𝑥 minus four because that was our 𝑑𝑢 𝑑𝑥.

Okay, great, but this isn’t a final answer because we can see there are actually still including 𝑢. So what we now need to do is actually substitute in our value for 𝑢. So therefore, we’re gonna get — and I’ve just rearranged a bit here — 18𝑥 minus four multiplied by four to the power of nine 𝑥 squared minus four 𝑥 plus eight because that was our 𝑢 and then that’s multiplied by the natural logarithm of four.

So therefore, after some tidying up using factoring, we can say that if 𝑦 is equal to four to the power of nine 𝑥 squared minus four 𝑥 plus eight, therefore 𝑑𝑦 𝑑𝑥 is gonna be equal to two multiplied by nine 𝑥 minus two. And we got that because actually we factored 18𝑥 minus four because two is a factor of 18𝑥 and negative four. And this is multiplied by four to the power of nine 𝑥 squared minus four 𝑥 plus eight ln four.