Video: Find the Derivative of a Parametric Equation

If, for 𝑑 > 0, π‘₯ = 𝑑³ and 𝑦 = sin (𝑑³), find d𝑦/dπ‘₯.

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Video Transcript

If, for 𝑑 is greater than zero, π‘₯ equals 𝑑 cubed and 𝑦 equals sin of 𝑑 cubed, find d𝑦 by dπ‘₯.

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

We can find dπ‘₯ by d𝑑 by differentiating 𝑑 cubed with respect to 𝑑. 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 d𝑑 equals three 𝑑 squared. So, now, we just need to find d𝑦 by d𝑑.

𝑦 equals sin of 𝑑 cubed, so we need to differentiate this with respect to 𝑑. But we actually have a function within a function here because we have the function 𝑑 cubed within the function sine. So, we differentiate this using the chain rule. We start by letting 𝑒 equal to our inner function 𝑑 cubed. Then we can say that 𝑦 equals sin of 𝑒. Then, by the chain rule, d𝑦 by d𝑑 equals d𝑦 by d𝑒 multiplied by d𝑒 by d𝑑.

Well, d𝑦 by d𝑒 is the derivative of sin of 𝑒 with respect to 𝑒. So, we recall that the derivative of sin of 𝑒 is cos of 𝑒. To get d𝑒 by d𝑑, we differentiate 𝑑 cubed with respect to 𝑑. Again, we can do this using the power rule, which gives us three 𝑑 squared. So, we found that d𝑦 by d𝑑 equals cos of 𝑒 multiplied by three 𝑑 squared.

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

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

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