27–40. Implicit differentiation Use implicit differentiat...

Question

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.

6x³+7y³ = 13xy

Accepted Answer

Hello. In this video, we are going to be finding the derivative DUIDX by using implicit differentiation. So, we want to implicitly differentiate the following equation. 5X4 minus 3 to 4th power is equal to 20XY. So how do we start the process of finding the derivative DY divided by DX? Well, in order to do this, we want to go ahead and differentiate with respect to the dependent variable. That variable is going to be X. So if we want to find the derivative DYDX, we will have to take the derivative of the entire equation with respect to X. So, what we'll need to do is we'll need to take the derivative of each term in the equation with respect to X. Let's go ahead and start by taking the derivative of the left hand side of the equation. Now, on the left-hand side, we have 5 X to the 4th power minus 3Y to the 4th power, and we are going to be taking the derivative with respect to X. So, the derivative of the first term, 5 X to the fourth power, will be 5 multiplied by 4 X to the 3rd power by the power rule. But what about the second term, 3 to the 4th power? Well, what we are going to do is we are going to treat the variable Y as a function. So we're going to take the derivative normally using the power rule. That derivative is going to be -3, multiplied by 4 to the third power, but because we're taking a derivative of Y with respect to X, through the rules of implicit differentiation, we will obtain a DYDX term. So, now let's go ahead and simplify the left-hand side. 5 multiplied by 4 will give us 20 X cubed, and -3 multiplied by 4 will give us -12y cubed DYDX. This is going to be the derivative of the left hand side. Now, what about the derivative of the right hand side? Well, for the right-hand side, we have 20 XY. And we are going to be taking the derivative with respect to X. Notice that we have a product of X and Y. So, in order to take this derivative, we are going to have to use the product rule. So, we will set 20 as the coefficient and multiply this by the derivative of X multiplied by Y. Through the product rule, that will give us X multiplied by the derivative of Y with respect to X. That derivative is going to be 1, and again, by implicit differentiation, we will need to generate a DYDX term. And one multiplied by DYDX will just be DYDX. Then we will add the second function Y multiplied by the derivative of X with respect to X, which is just 1, and 1 multiplied by Y is just Y. Now let's simplify the right-hand side. We will distribute 20 to XDYDX and to Y, and that will give us 20 XDYDX. Plus 2 Y. Now that we have the derivatives of the left hand side and right hand side of the equation, let's go ahead and plug this back into an equation form. So that is going to be 20 X cubed. Minus 12 Y cubed DYDX equal to 20 XDYDX. Plus 2 Y. Now, again, going back to the beginning of the problem, our goal is to solve for the derivative DYDX. So what we need to do now is we need to solve for the DYDX terms in our current derivative. In order to do this, what we are going to do is we are going to move all the DYDX terms to one side of the equation, and all the constant terms to the other side. So here, we're going to add 12 Y cubed DYDX to both sides of the equation. And we will subtract 20 Y to both sides of the equation. That will allow us to rewrite the derivative as 20 x cubed minus 20 Y equal to 20XDYDX plus 12 Y cubed DYDX. Now, notice that all the terms on the right-hand side have a factor of DYDX. Now, what we can go ahead and do is factor out a DYDX from the right-hand side of the equation. That is going to leave us with 20 X cubed minus 20 Y equal to the quantity 20 X plus 12y cubed. Multiplied by DYDX. And finally, in order to solve for DYDX, we will divide both sides of the equation by 20 X plus 12 Y cubed. That will give us DYDX equal to the quantity 20 X cubed minus 20 Y divided by 20 X plus 12 Y cubed. Now, one more thing that we can do in order to simplify this derivative is we can factor out a 20 from the numerator and factor out. A 4 from the denominator. That will allow us to simplify the derivative to be 20 multiplied by X cubed minus Y. Divided by 4 multiplied by 5 X. Plus 3 Y cubed. Finally, we can simplify the constants 20 and 4, and that will simplify the derivative to be DYDX equal to 5 multiplied by X cubed minus Y. Divided by 5 X plus 3 Y cubed, and this is going to be the derivative DYDX by using implicit differentiation. And with that being said, the solution to the problem is going to be C. So, I hope this video helps you in understanding how to use implicit differentiation, and I will go ahead and see you all in the next video.

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.
6x³+7y³ = 13xy

Key Concepts

Implicit Differentiation

Chain Rule

Derivative Notation (dy/dx)

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