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.

x+y³−xy=1 (Hint: Rewrite as y³−1=xy−x and then factor both sides.)

Accepted Answer

Hello. In this video, we are going to be approaching the following derivative problem. We want to determine the derivative DY divided by DX using implicit differentiation for the equation 4 X plus Y cubed minus 2 XY is equal to 8. We want to rearrange the terms of this equation as follows, and then factor both sides. So, this problem is telling us to use the form of the equation Y cubed minus 8 is equal to 2 XY minus 4 X. And the first thing we want to do is we want to go ahead and factor both sides of this equation. For the left hand side, notice that we can factor this equation by using the difference of cubes. By using the difference of cubes, we can factor the left hand side of the equation to be Y minus 2 multiplied by the quantity Y2 + 2 Y + 4. Now, what about the right hand side? Well, notice that on the right-hand side, each term has a factor of 2 X. So we can factor out a 2 X from the right hand side of the equation, allowing us to rewrite it as Y minus 2X multiplied by the quantity Y minus 2. Now, at this step, we can go ahead and divide both sides of the equation by Y minus 2, but this produces two situations that we need to consider. First, if we divide by Y minus 2, then Y minus 2, or we are going to divide under the assumption that Y minus 2 is not equal to 0. So let's go ahead and solve for the derivative with the assumption that Y minus 2 is not equal to 0. So, when we divide both sides of the equation by that quantity, the equation will simplify to be Y2 + 2 Y + 4 equal to 2 X. Now what we want to do is we want to go ahead and take the derivative of both sides of the equation with respect to X. So, let's go ahead and take the derivative of Y2d. By the power rule, that derivative is going to be 2 Y. But since we are taking the derivative of Y with respect to X, we are going to generate a DYDX term through implicit differentiation. Next, the derivative of 2 Y is just going to be 2 by the power rule, and through implicit differentiation, we get a DYDX term. Now, the derivative of 4 is 0, since the derivative of a constant value is always 0. Now, on the right-hand side, the derivative of 2 X is going to be 2 through the power rule, and we will not get a DYDX term since we are taking the derivative of X with respect to X. Now what we are going to do is we are going to factor out the derivative DYDX from both sides of the equation. That will give us D Y D X multiplied by the quantity 2 Y plus 2 equal to 2. And if we divide both sides of this equation by 2Y + 2, we can rewrite the derivative to be DYDX equal to 2 divided by 2y plus 2. We can further factor a 2 from the denominator, leaving us with 2 divided by 2 multiplied by Y + 1, and if we reduce the factors of 2 in both the numerator and denominator, the simplest form of this derivative is going to be Y divided by Y + 1. So this is going to be the derivative DYDX under the assumption that Y minus 2 is not zero. But now we need to consider the other half of the case, where the quantity Y minus 2 is equal to 0. Well, in the case that Y minus 2 is equal to 0, if we solve for Y in this case, we get the equation Y is equal to 2. Then if we were to take the derivative of both sides of this equation with respect to X, that is going to give us DYDX equal to 0, and that is going to be the derivative with respect to X under the assumption that Y minus 2 is equal to 0. And with that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding on how to approach this type of problem, and I will go ahead and see you all in the next video.

The following equations implicitly define one or more functions.a. Find dy/dx using implicit differentiation.x+y³−xy=1 (Hint: Rewrite as y³−1=xy−x and then factor both sides.)

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The following equations implicitly define one or more functions.
a. Find dy/dx using implicit differentiation.
x+y³−xy=1 (Hint: Rewrite as y³−1=xy−x and then factor both sides.)