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

Suppose π₯ = 9π β 5 and π¦ = 4β(π) + 6πΒ². Find dy/dπ₯ when π = 4.

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### Video Transcript

Suppose π₯ equals nine π minus five and π¦ equals four root π plus six π squared. Find dπ¦ by dπ₯ when π is equal to four.

Here we have a pair of parametric equations expressed in terms of a third parameter, π. We recall that, for two differentiable functions, π and π, where π₯ is equal to π of π‘ and π¦ is equal to π of π‘, dπ¦ by dπ₯ is found by dividing dπ¦ by dπ‘ by dπ₯ by dπ‘.

Now of course, our functions are in terms of π. So here we can say that dπ¦ by dπ₯ must be equal to dπ¦ by dπ over dπ₯ by dπ. And then itβs quite clear weβre going to need to differentiate both of our functions with respect to π. Well, the derivative of nine π is nine. And the derivative of negative five is zero. So dπ₯ by dπ is nine.

Letβs write four root π as four times π to the power of one-half. Then the derivative of four π to the power of one-half is a half times four π to the power of negative one-half, which simplifies to two π to the power of negative one-half. Then the derivative of six π squared is two times six π, which is 12π. dπ¦ by dπ₯ is then the quotient of these. Itβs two π to the power of negative one-half plus 12π all over nine.

Now of course, weβre looking to find the value of the derivative when π is equal to four. So letβs substitute π equals four into our function. Thatβ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 π is equal to four, dπ¦ by dπ₯ is equal to 49 over nine.