The following equations implicitly define one or more function...

Question

The following equations implicitly define one or more functions.

a. Find dy/dx using implicit differentiation.

y² = x²(4 − x) / 4 + x (right strophoid)

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the derivative DUIDX using implicit differentiation for the equation Y2 E equals 3x2 multiplied by 5 minus X, all divided by 6 + 2X. Now if we're going to use implicit differentiation to solve for DUID X, we're going to have to differentiate both sides of our equation with respect. The X. Now if we start with our left hand side, let me actually write what we're doing here. So we're differentiating both sides with respect to X. And I need to spell implicitly properly here. Implicitly, obviously I have a problem with spelling. Implicitly with respect to X, OK. Starting with our left hand side, I'll put that in blue. Then we're going to be finding the derivative with respect to X of Y squared. Which is gonna be equal to 2 Y D YD X. Now our right hand side seems a little bit more complicated because we have a cautient here. So that means we're going to differentiate the right hand side with the cautient rule, OK? Now, can you recall what the quotient rule tells us? Well, recall. OK, that if we have a function Y that's equal to you divided by V a quotient. Then the derivative of y with respect to X is going to be equal to Vu minusuv divided by V2 where U and V represent the derivatives of U and V respectively. So, let's set you, OK, to the value of our numerator, 3 X squared multiplied by 5 minus X. And let's say V to be. 6 + 2 X or denominator. We're going to differentiate both of those, plug them into the quotient rule to differentiate with respect to X, and then put our answer back into the implicit derivative of our equation, OK? Now let's start with you. If we expand U, it's gonna become 15 X2 minus 3XQ and thus the derivative of U is going to be equal to 30 X minus 9 X 2 by the power rule. On the other hand, for our denominator, the derivative of 6 + 2 X is 2. The derivative of 6 is 0 and the derivative of 2 X is 2. So now, what this tells us then. Is that for our right hand side, it's derivative, OK, is gonna be equal to, so the derivative of 3 X squared multiplied by 5 minus X divided by 6 + 2 X. Is going to be equal to remember V multiplied by U. OK, so that's 6 + 2 X. Multiplied by 30 X minus 9 X squared. Minus you multiplied by a V, so that's 15 X squared. OK, already established that. 3 XQed multiplied by 2. Now all of this is being divided by 6 + 2 X2. Now, can we simplify this a bit further before we put it back into our equation? Well, here We can do some expanding, right? Cause if we expand our term here, 6 + 2 X multiplied by 30X minus 9X squared, it's gonna become 180X minus 64 X squared. And I remember we're multiplying the whole thing. OK. Plus 60 x 2 minus 18 XQ. And on the other hand, for 15 X2 minus 3Xcube multiplied by 2, it's gonna become 30 X2 minus 6 X cubed, OK? And now if we get rid of our brackets for that one, we're gonna have minus 30 X2 plus. Plus 6 X cubed, OK? And all of this again will be under RB over 6 + 2 X squared. And if we simplify your numerator here, OK, then we're gonna have 180 X. OK, minus 24 X squared. OK. Minus 12 XQ. Know that we have the derivative on our right hand side and our left hand side, we can put them back into our equation. So no, that means then we're gonna have 2 Y DYDX. Being equal to. 180 X minus 24 X2 minus 12 X cubed. OK, divided by 6 + 2 X all squared. But where do we go from here? Well, we can divide through our equation by 2y. OK. To get this term, I'm just gonna copy it in the interest of time here. And we're gonna be dividing that by 2 Y, OK. So we'll have our 2 Y here. Now if we think about it, we could do some, we could, we could simplify this even further. Also, well, in our numerator, OK, we could factor out 12 X. So that's gonna be 12 X multiplied by 15 minus 2X minus X squared. And in our denominator we could factor out uh 2 from 6 + 2 X. OK. And we know it's gonna be squared. So if we factor it out again, then we're gonna have 8 Y multiplied by 3 + X all squared. Now, if we take this a bit further here, OK, let me make some space. If we, if we take this a bit further here, then we're gonna have, we can factor out 4, to get 3 X multiplied by 150 minus 2X minus X squared. OK, all divided by 2 Y multiplied by 3 + X squared. OK. If I make some space here. Then we could even go further to factor 15 minus 2X minus X squared, OK, do some quadratic factorization here to get DUID X. To be equal to -3 x multiplied by x minus 3 multiplied by x + 5 divided by 2y multiplied by 3 + X2 thus by implicitly differentiating this is going to be the derivative of Y with respect to X. Thanks a lot for watching everyone. I hope this video helped.

The following equations implicitly define one or more functions.
a. Find dy/dx using implicit differentiation.
y² = x²(4 − x) / 4 + x (right strophoid)

Key Concepts

Implicit Differentiation

Chain Rule

Right Strophoid

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