66–71. Higher-order derivatives Find and simplify y''.

Question

x + sin y = y

Accepted Answer

Welcome back, everyone. In this problem, we want to find the second derivative of the equation E X + Y equals 2 Y. A says it's e x multiplied by 2 + sy squared plus e xy all divided by 2 + y. B E X multiplied by 2 + sy plus e 2 x y divided by 2 + sy. C E X multiplied by 2 + sysquared minus e x y divided by 2 + siny cubed. And D E X multiplied by 2 + siney plus e x y divided by 2 + y cubed. Now what do we know about the second derivative? Well, basically that just means we're going to differentiate Y twice. So let's see if we can find the 1st 1st derivative of Y and then differentiate it again to find the second derivative. Now here, notice that we're not given Y as an equation or a function directly, OK, because our equation has Y on both sides and we're finding the cosine of Y. So Y is an argument of the cosine function as well. So we won't be able to solve for Y, but instead we can differentiate implicitly. That is, we can take the derivative of each side. So when we do that, then the derivative of the, let me write that here, let me put that in red. So differentiating implicitly. To solve for the first derivative of Y. OK. Then to differentiate both sides, on the left, we're gonna have the derivative of E to the X with respect to X. Plus the derivative of the cosine of y with respect to X equal to the derivative of 2y with respect to X. Now what's what's gonna be our results? Well, here the derivative of e X with respect to X is equal to e X. The derivative of the cosine of X is negative sine X. So since this is a composite function here, we can use the chain rule. To get the derivative of a cosiney with respect to X to be equal to negative sine Y multiplied by the derivative of Y, OK, with respect to X. And on our right hand side, this will be 2 multiplied by the derivative of Y with respect to X. No, we can try to isolate to get the first derivative, OK? Because no, that means we could rewrite this as E to the X being equal to 2 minus, plus, sorry, sine multiple. You write it in a different way here. Multiply plus Y multiplied by sin Y. OK. And now we can isolate Y to get Y multiplied by 2 plus siny. And now in this case, we can solve for the first derivative because it's gonna be equal to E X divided by 2 plus sine Y. OK? So now we have our first derivative. Next, we need to solve for our 2nd derivative. Since we have our first derivative and we differentiate it to get the second derivative, then. It means we're gonna be differentiating Y. Now take a good look at Y. It's a quotient. That means to differentiate it, we can use the quotient rule, OK? Now, what does the quotient rule of differentiation say? Well, basically what the quotient rule tells us, uh let me write it in a box over here, OK. It basically tells us that if Y is equal to u divided by V where U and V are differentiable functions, then the derivative of Y is going to be equal to vu minusuv divided by V2d, where U and V are the derivatives of U and V respectively. That is the quotient rule. Now if we look at our expression for Y, notice that our numerator is eX, so we can call that U, and our denominator is 2 + siney. Let's call that V. So we're going to differentiate our numerator and denominator, and then we're going to plug them into the quotient rule to get the second derivative. So now when we let you be equal to E to the X. Then the derivative of you is going to be eaten at X. Next, if we let V be equal to 2 + siney, OK. Then the derivative of V is going to be equal to 0. The constant, sorry, the derivative of constant is 0, plus YY. So the derivative of V is YY. Now that we have UV and their derivatives, we can plug them into the quotient rule because no, that means the second derivative then. Is going to be equal to, like we said, Zu. OK, so no, that's gonna be E to the X. Multiplied by 2 plus sine y. Minus UV. OK, UV. So that's E to the X multiplied by Y + Y. I know all of this is being divided by V squared, so 2 plus siney. Squirt Now here, notice that in our numerator both terms have a common factor of e to the X, so we can factor E to X out. Now, as we're writing, OK, notice that here we're gonna have minus Y + Y when you factor E to the X, but we already know what Y is. So we can replace Y here with E X divided by 2 + siny. And write it being multiplied by the cosine of Y. No, in this case, notice that in our denominator we also have the expression 2 plus sine y squared, so we could try to write our expression in the numerator with 1 as one fraction, and then we could simplify it even further because no under a single fraction, OK. Then in our numerator. We're going to have 2 + sine Y multiplied by 2 + sine Y. That's 2 + Y squared, OK. Minus our terms here e to the X multiplied by the cosine of Y, OK. I know all of this is going to be divided by 2 + sine squared multiplied by 2 + siney, which is gonna be 2 + siny or cued. Thus, this is the second derivative of Y. If we look back at our answer choices, that tells us then that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

66–71. Higher-order derivatives Find and simplify y''.

x + sin y = y

Key Concepts

Implicit Differentiation

Higher-Order Derivatives

Chain Rule

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