# Video: Find the Derivative of a Parametric Equation

If, for π‘ > 0, π₯ = π‘Β³ and π¦ = sin (π‘Β³), find dπ¦/dπ₯.

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### 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. Itβs 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, weβre 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.