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

Find the first derivative of the function π¦ = ln (β5π₯β΄ + 2π₯Β²).

03:28

### Video Transcript

Find the first derivative of the function π¦ equals ln of negative five π₯ to the power of four plus two π₯ squared.

So this is the function weβve been asked to differentiate, which is in fact a function of a function. And the way to differentiate a function of a function is by using the chain rule. The chain rule says that if π¦ equals π of π’ and π’ equals π of π₯ then dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯. And for our question, π’ is the inner function negative five π₯ to the power of four plus two π₯ squared. And π¦ is the outer function ln π’.

We can see from the formula for the chain rule that weβre going to need dπ¦ by dπ’ and dπ’ by dπ₯. So, letβs find dπ¦ by dπ’ first. Well, π¦ equals ln of π’. So, weβre gonna differentiate this with respect to π’. And to do this, we recall the general rule that tells us if π¦ equals ln of π₯, then dπ¦ by dπ₯ equals one over π₯. And so for π¦ equals ln of π’, dπ¦ by dπ’ equals one over π’.

Okay, so now, we need to find dπ’ by dπ₯. And remember that we have that π’ is equal to negative five π₯ to the power of four plus two π₯ squared. And before we differentiate this, letβs just remember the power rule of differentiating. That is, if π¦ equals ππ₯ to the power of π, dπ¦ by dπ₯ equals πππ₯ to the power of π minus one. And so, with this in mind, dπ’ by dπ₯ is equal to negative 20π₯ to the power of three plus four π₯.

And so now, applying the formula for the chain rule, dπ¦ by dπ₯ equals one over π’ multiplied by negative 20π₯ to the power of three plus four π₯, which we can write as a single fraction. And we also remember that we already defined π’ as negative five π₯ to the power of four plus two π₯ squared.

And to tidy this up, since we have a negative here and here, weβll multiply by negative one over negative one to get 20π₯ to the power of three minus four π₯ over five π₯ to the power of four minus two π₯ squared. And we can actually take out π₯ as a common factor in both the numerator and the denominator. Which means we can cancel π₯ on the top and bottom. And that gives us 20π₯ squared minus four over five π₯ to the power of three minus two π₯. And for our final answer, weβll take out π₯ as a common factor on the denominator. And thereβs our final answer.