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(a) Find $ y' $ by implicit differentiation. (b)…

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Problem 1 Easy Difficulty

(a) Find $ y' $ by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get $ y' $ in terms of $ x. $
(c) Check that your solutions to part (a) and (b) are consistent by substituting the expression for $ y $ into your solution for part (a).

$ 9 x^2 - y^2 = 1 $

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Video Transcript

Alright, here's a cool problem. We have a function. Nine X squared minus Y squared equals one. And we are going to find our derivative by implicit differentiation. So let's write that down, implicit differentiation. Alright, okay. So that means we can just start taking the derivative. We don't have to solve for Y. First. And so when we take the derivative we get we'll do power roles. So we'll get nine times two. So 18 X to the first power minus. We're going to do power rule and chain role. So two Y to the first power D. Y. D. S And derivative of one is 0. I want to get dy dx by itself. So I'm going to go ahead and Um I'm going to subtract 18 X from both sides and then divide by minus two. Let me do it in two steps just so it's clear to everyone. So let's go back and actually do it um in two steps and I'll actually just do it below and be a little easier to see what's happening. Okay, so first thing I'm gonna do is subtract um and I don't need to put first power. Okay, so I'm going to subtract 18 acts from both sides so we can see every step clearly And then I'm gonna divide both sides by -2 Y. And that will get dy dx by itself. I can clean it up just a tad And that will give me nine x over Y. Alright, so there is the first part solved we found um we use implicit differentiation to find the derivative of ry function. All right, excellent. Alright let's do part B. Part B. Now says let's solve for y first and then take the drift of that way. Alright, so this is our equation. I'll rewrite it nine X squared minus Y squared equals one. I'm going to subtract nine X squared from both sides And then multiply by -1 both sides. And that will give me I can reverse the order here then we square root both sides. But that will give me plus or minus square root of nine X squared minus one. And then I can write as nine X squared minus one to the one half Power. Alright now we're going to take the derivative directly. So we'll use power rule chain rule. So dy let's change color here. Just for emphasis that we're now doing the derivative. Okay, so dy dx it's going to be plus or minus. We bring the one half down and we keep the in inner argument the same. But bring the power down to subtracting one from a half which is minus a half. And then we have chain rules. So we multiply by the derivative of the inside which is 18 X. Alright, so this looks we can clean it up a tag. Let's see we're gonna get half of 18 is nine. We still have plus or minus. So on the top is nine X. On the bottom is square root of nine X squared minus one and uh Oh cool. Look at that square, whoops go one lower. Plus or minus. Look at this plus or minus uh square root of nine X squared minus one equals to y. So I can replace this by nine X over Y. And um we get the same result as we did for implicit. So very fun, very cool. So it works that they that the two methods work there the same. Alright, hopefully that helped to have a wonderful day.

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Calculus: Early Transcendentals

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