Video: Finding the First Derivative of a Function Defined by Parametric Equations at a Given Value for the Parameter

Video Transcript

Suppose 𝑥 equals nine 𝑛 minus five and 𝑦 equals four root 𝑛 plus six 𝑛 squared. Find d𝑦 by d𝑥 when 𝑛 is equal to four.

Here we have a pair of parametric equations expressed in terms of a third parameter, 𝑛. We recall that, for two differentiable functions, 𝑓 and 𝑔, where 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡, d𝑦 by d𝑥 is found by dividing d𝑦 by d𝑡 by d𝑥 by d𝑡.

Now of course, our functions are in terms of 𝑛. So here we can say that d𝑦 by d𝑥 must be equal to d𝑦 by d𝑛 over d𝑥 by d𝑛. And then it’s quite clear we’re going to need to differentiate both of our functions with respect to 𝑛. Well, the derivative of nine 𝑛 is nine. And the derivative of negative five is zero. So d𝑥 by d𝑛 is nine.

Let’s write four root 𝑛 as four times 𝑛 to the power of one-half. Then the derivative of four 𝑛 to the power of one-half is a half times four 𝑛 to the power of negative one-half, which simplifies to two 𝑛 to the power of negative one-half. Then the derivative of six 𝑛 squared is two times six 𝑛, which is 12𝑛. d𝑦 by d𝑥 is then the quotient of these. It’s two 𝑛 to the power of negative one-half plus 12𝑛 all over nine.

Now of course, we’re looking to find the value of the derivative when 𝑛 is equal to four. So let’s substitute 𝑛 equals four into our function. That’s two times four to the power of negative one-half plus 12 times four over nine. Well, four to the power of negative one-half is one over the square root of four, or one-half. And two times one-half is just [one]. We know that 12 times four is 48 and one plus 48 is 49. And so when 𝑛 is equal to four, d𝑦 by d𝑥 is equal to 49 over nine.