Question Video: Differentiating Logarithmic Functions Using the Chain Rule | Nagwa Question Video: Differentiating Logarithmic Functions Using the Chain Rule | Nagwa

# Question Video: Differentiating Logarithmic Functions Using the Chain Rule Mathematics

Find ππ¦/ππ₯, given that π¦ = ln (4π₯ + 5)β·.

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### Video Transcript

Find ππ¦ ππ₯, given that π¦ equals the natural logarithm of four π₯ plus five to the power of seven.

So the first thing that weβ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 weβve got log to the base π of π₯ to the power of π, then this can be rewritten as π log to base π of π₯. So when we rewrite our function, we actually get π¦ is equal to seven ln four π₯ plus five. So we did that using the log law.

So now if weβre actually looking to find ππ¦ ππ₯, we can say itβs gonna be equal to seven multiplied by πππ₯ of ln four π₯ plus five. So what we wanna do now is actually differentiate our ln four π₯ plus five. And the way to do that is actually by using the chain rule. And the chain rule actually states that ππ¦ ππ₯ is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. So we can actually adapt this and use this to solve our problem.

So if weβre looking to differentiate ln four π₯ plus five, we can actually say that π’ is gonna be equal to four π₯ plus five, which is the actual expression inside the parentheses, and π¦ is gonna be equal to ln π’. So first of all, what weβre actually gonna do is differentiate π’. So itβs ππ’ ππ₯, which is just gonna give us four, because if we differentiate four π₯, we get four. And if we differentiate positive five, it actually just turns to zero. So great! ππ’ ππ₯ is equal to four.

So now weβre gonna find ππ¦ ππ’. And to do that, weβre actually gonna differentiate ln π’. Weβre going to get one over π’. And we get that because actually thereβs a general form that tells us that if π¦ is equal to ln π₯, then ππ¦ ππ₯ is equal to one over π₯.

Okay, great! So now weβve actually differentiated both parts. We can use the chain rule to put it together. So weβre gonna get that the differential of ln four π₯ plus five is equal to four, because that was ππ’ ππ₯ multiplied by one over π’, which is ππ¦ ππ’. So now just substitute in our value for π’, which gives us four over four π₯ plus five.

Okay, great! So now weβve actually differentiated our ln four π₯ plus five. Thereβs just one more step to do to actually find our final ππ¦ ππ₯. So weβre gonna get ππ¦ ππ₯ is equal to seven multiplied by four over four π₯ plus five. And thatβs, as we said before, was because we actually differentiated ln four π₯ plus five. So therefore we can say that, given that π¦ equals ln four π₯ plus five to the power of seven, ππ¦ ππ₯ is equal to 28 over four π₯ plus five.

Itβs worth mentioning at this point that we couldβve actually used the formula to help us differentiate ln four π₯ plus five. And that would have been that if it was in the form π¦ equals ln π π₯, then ππ¦ ππ₯ is equal to the differential of π π₯ over π π₯. 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 π π₯, was actually four π₯ 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 wouldβve been the function itself, which was four π₯ 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 π₯ plus five using the chain rule.