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

Suppose π‘₯ = 9𝑛 βˆ’ 5 and 𝑦 = 4√(𝑛) + 6𝑛². Find dy/dπ‘₯ when 𝑛 = 4.

01:45

Video Transcript

Suppose π‘₯ equals nine 𝑛 minus five and 𝑦 equals four root 𝑛 plus six 𝑛 squared. Find d𝑦 by dπ‘₯ 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 π‘₯ is equal to 𝑓 of 𝑑 and 𝑦 is equal to 𝑔 of 𝑑, d𝑦 by dπ‘₯ is found by dividing d𝑦 by d𝑑 by dπ‘₯ by d𝑑.

Now of course, our functions are in terms of 𝑛. So here we can say that d𝑦 by dπ‘₯ must be equal to d𝑦 by d𝑛 over dπ‘₯ 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 dπ‘₯ 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 dπ‘₯ 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 dπ‘₯ is equal to 49 over nine.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.