Video: Differentiating Logarithmic Functions Using the Chain Rule

Video Transcript

Find 𝑑𝑦 𝑑𝑥, given that 𝑦 equals the natural logarithm of four 𝑥 plus five to the power of seven.

So the first thing that we’re looking to do when we differentiate our function is actually rewrite it in another way. And we can do that because of one of the log laws. And that law states that if we’ve got log to the base 𝑏 of 𝑥 to the power of 𝑛, then this can be rewritten as 𝑛 log to base 𝑏 of 𝑥. So when we rewrite our function, we actually get 𝑦 is equal to seven ln four 𝑥 plus five. So we did that using the log law.

So now if we’re actually looking to find 𝑑𝑦 𝑑𝑥, we can say it’s gonna be equal to seven multiplied by 𝑑𝑑𝑥 of ln four 𝑥 plus five. So what we wanna do now is actually differentiate our ln four 𝑥 plus five. And the way to do that is actually by using the chain rule. And the chain rule actually states that 𝑑𝑦 𝑑𝑥 is equal to 𝑑𝑦 𝑑𝑢 multiplied by 𝑑𝑢 𝑑𝑥. So we can actually adapt this and use this to solve our problem.

So if we’re looking to differentiate ln four 𝑥 plus five, we can actually say that 𝑢 is gonna be equal to four 𝑥 plus five, which is the actual expression inside the parentheses, and 𝑦 is gonna be equal to ln 𝑢. So first of all, what we’re actually gonna do is differentiate 𝑢. So it’s 𝑑𝑢 𝑑𝑥, which is just gonna give us four, because if we differentiate four 𝑥, we get four. And if we differentiate positive five, it actually just turns to zero. So great! 𝑑𝑢 𝑑𝑥 is equal to four.

So now we’re gonna find 𝑑𝑦 𝑑𝑢. And to do that, we’re actually gonna differentiate ln 𝑢. We’re going to get one over 𝑢. And we get that because actually there’s a general form that tells us that if 𝑦 is equal to ln 𝑥, then 𝑑𝑦 𝑑𝑥 is equal to one over 𝑥.

Okay, great! So now we’ve actually differentiated both parts. We can use the chain rule to put it together. So we’re gonna get that the differential of ln four 𝑥 plus five is equal to four, because that was 𝑑𝑢 𝑑𝑥 multiplied by one over 𝑢, which is 𝑑𝑦 𝑑𝑢. So now just substitute in our value for 𝑢, which gives us four over four 𝑥 plus five.

Okay, great! So now we’ve actually differentiated our ln four 𝑥 plus five. There’s just one more step to do to actually find our final 𝑑𝑦 𝑑𝑥. So we’re gonna get 𝑑𝑦 𝑑𝑥 is equal to seven multiplied by four over four 𝑥 plus five. And that’s, as we said before, was because we actually differentiated ln four 𝑥 plus five. So therefore we can say that, given that 𝑦 equals ln four 𝑥 plus five to the power of seven, 𝑑𝑦 𝑑𝑥 is equal to 28 over four 𝑥 plus five.

It’s worth mentioning at this point that we could’ve actually used the formula to help us differentiate ln four 𝑥 plus five. And that would have been that if it was in the form 𝑦 equals ln 𝑓 𝑥, then 𝑑𝑦 𝑑𝑥 is equal to the differential of 𝑓 𝑥 over 𝑓 𝑥. However, I just wanted to demonstrate how the chain rule is used to actually get to this result. And we can actually double-check it using this, because if we actually look, our function, so our 𝑓 𝑥, was actually four 𝑥 plus five.

So, therefore, if we differentiate this, we just get four. So that would have been our numerator. And then on the denominator, it just would’ve been the function itself, which was four 𝑥 plus five. And as we can see in the previous step, just before the final answer, this was in fact the value that we found when we differentiated ln four 𝑥 plus five using the chain rule.