### Video Transcript

If, for π‘ is greater than zero, π₯ equals π‘ cubed and π¦ equals sin of π‘ cubed, find dπ¦ by dπ₯.

We can see here that π₯ and π¦ are defined in terms of another variable, π‘. So, we call these parametric equations. And the rule to find dπ¦ by dπ₯ when π₯ and π¦ are defined in terms of π‘ parametrically. Itβs just an application of the chain rule. The chain rule tells us that dπ¦ by dπ‘ equals dπ¦ by dπ₯ multiplied by dπ₯ by dπ‘. And so, by rearrangement, we have that dπ¦ by dπ₯ equals dπ¦ by dπ‘ over dπ₯ by dπ‘. But of course, for this to work, we must have that the denominator, dπ₯ by dπ‘, is not equal to zero. So, weβre going to need to find dπ¦ by dπ‘ and dπ₯ by dπ‘.

We can find dπ₯ by dπ‘ by differentiating π‘ cubed with respect to π‘. We can do this using the power rule, which tells us to multiply by the power and then subtract one from the power. And so, we find that dπ₯ by dπ‘ equals three π‘ squared. So, now, we just need to find dπ¦ by dπ‘.

π¦ equals sin of π‘ cubed, so we need to differentiate this with respect to π‘. But we actually have a function within a function here because we have the function π‘ cubed within the function sine. So, we differentiate this using the chain rule. We start by letting π’ equal to our inner function π‘ cubed. Then we can say that π¦ equals sin of π’. Then, by the chain rule, dπ¦ by dπ‘ equals dπ¦ by dπ’ multiplied by dπ’ by dπ‘.

Well, dπ¦ by dπ’ is the derivative of sin of π’ with respect to π’. So, we recall that the derivative of sin of π’ is cos of π’. To get dπ’ by dπ‘, we differentiate π‘ cubed with respect to π‘. Again, we can do this using the power rule, which gives us three π‘ squared. So, we found that dπ¦ by dπ‘ equals cos of π’ multiplied by three π‘ squared.

But we remember that we let π’ equal to π‘ cubed. So, we can replace π’ with π‘ cubed. So, now, we have dπ₯ by dπ‘ and dπ¦ by dπ‘, we can find dπ¦ by dπ₯. We have that dπ¦ by dπ₯ equals dπ¦ by dπ‘, which is cos of π‘ cubed multiplied by three π‘ squared over dπ₯ by dπ‘, which is three π‘ squared.

Remember, the condition for this is that dπ₯ by dπ‘ must not be equal to zero. But as we were told in the question that π‘ is always bigger than zero, then three π‘ squared must also be bigger than zero. So, that condition is satisfied. From here, we can cancel the three π‘ squared. And this gives us cos of π‘ cubed.