### Video Transcript

Find dπ¦ dπ₯, given that π¦ is equal to three π₯ to the power of six ln three π₯ cubed.

So if weβre looking to differentiate this function, the first thing we can do is take the constant value, which is our three at the beginning, outside and then deal that afterwards because that wonβt affect the differentiation. So therefore, we can say the derivative of π¦ is equal to three multiplied by the derivative of π₯ to the power of six ln three π₯ to the power of three. So what we can see now is that weβve got π₯ to the power of six multiplied by ln three π₯ to the power of three. So we can use the product rule because itβs like a function multiplied by another function.

And the product rule tells us that if we have π of π₯ multiplied by π of π₯, then if we say that π’ is equal to π of π₯ and π£ is equal to π of π₯, then derivative of π’π£ is equal to π’ dπ£ dπ₯ plus π£ dπ’ dπ₯ which is our first term multiplied by the derivative of the second term plus our second term multiplied by the derivative of the first term. Okay great, now, we have that; we can apply that to solve our problem.

So first of all, we can say the π’ is equal to π₯ to the power of six. And therefore if we differentiate this, weβre gonna get six π₯ to the power of five. Thatβs because what we do is we multiply the exponent by the coefficient, so six by one which gives us six. And then we reduce the exponent by one. So we go from six to five. So we got six π₯ to the power of five. And then next, we can say that π£ is equal to ln three π₯ cubed.

And to be able to differentiate this, what we can use is an adaptation of the chain rule. And this tells us that the derivative of ln π’ of π₯ is equal to one over π’ of π₯ multiplied by the derivative of π’ of π₯. And in our expression, the three π₯ cubed is our π’ of π₯. So therefore, we can say that dπ£ dπ₯ is gonna be equal to one over three π₯ cubed multiplied by β β then weβve got nine π₯ squared, and thatβs because this is the derivative of three π₯ cubed. So again, exponent multiplied by coefficient is three multiplied by three which is nine. Then we reduce the exponent by one to two.

Okay great, we can now simplify this. And to first simplify, weβre gonna divide the top by three and the bottom by three. So weβre gonna get three π₯ squared. And weβre gonna have π₯ cubed on the bottom. And then we can divide three by π₯ squared. So then weβre left with one multiplied by three on the numerator and just π₯ in the denominator. So weβre left with dπ£ dπ₯ is equal to three over π₯. So now, itβs time to apply our product rule and bring everything back together. So first fist of all, weβre gonna have three multiplied by β then weβre gonna have three π₯ to the power of six over π₯. And thatβs because thatβs our π’dπ£ dπ₯ because π’ was π₯ to the power of six. And dπ£ dπ₯ was three over π₯. And this is gonna be plus and then three π₯ cubed multiplied by six π₯ to the power of five. And this is because this is our π£dπ’ dπ₯.

Well now, we can look to simplify. So first of all, in the first term, what we can do is divide the top and bottom or numerator and denominator by π₯. So weβre left with three π₯ to the power of five. So then what we can do is we can distribute the three across the parentheses. And when we do this we get nine π₯ to the power of five plus 18π₯ to the power of five ln three π₯ cubed. So therefore, we can say that, given that π¦ is equal to three π₯ to the power of six ln three π₯ cubed, then dπ¦ dπ₯ is equal to nine π₯ to the power of five plus 18π₯ to the power of five ln three π₯ cubed.