In Exercises 53 and 54, find both dy/dx (treating y as a diffe...

Question

In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?

53. xy³ + x²y = 6

Accepted Answer

Welcome back, everyone. Given the equation 2 X Y minus 3 X Y 2 equals 4, find both D Y divided by DX and D X divided by D Y. Treading Y as a function of X and vice versa. What is the relationship between DY divided by DX? And the X divided by D Y. So first of all, let's identify the Y divided by D X, which means that Y is a function of X. This means that we want to differentiate to X 2 Y. Subtract 3 multiplied by x y2 derivative. And set it equals to the derivative of 4. Let's identify all of the three derivatives. We can begin with the derivative of X Y. And we can apply the product rule. This means that we're differentiating X2. Multiplied by Y plus x2 multiplied by Y derivative. And according to the power rule, this is 2 XY. Plus x2y. Similarly, let's identify X Y2 derivative, which is the derivative of X multiplied by Y2. Plus X multiplied by Y2 derivative. And this is going to be equal to Y2 because the derivative of X is 1. Plus 2 X. Notice that the derivative of Y2 is 2 Y, applying the power rule, and according to the chain rule we're also multiplying by the derivative of Y because Y is a function of X. So now what we're going to do is basically substitute those in. We have 2 multiplied by. 2 XY, which is 4 XY. Plus 2 X2 minus 3 Y2. Minus 6 Y. Y X and that must be equal to 0 because the derivative of a constant is 0. Now, let's factor out the term. Which leaves us with 2 X squares. Minus 6 X. Why And this must be equal to. Wey we're moving the negative terms to the right. And minus. 4 XY once again were moving the other term to the right. This allows us to solve our Y as. 3 y2 minus 4 XY. Divided by 2 X2 minus 6 XY. So we have our first derivative. Which is DY divided by D X. Let's label it as our first answer. Now, for the second part of this problem, we want to identify D X divided by D Y. So once again, we're going to differentiate the same expression. 2 multiplied by the derivative of X2 y. -3 multiplied by X Y. Squared derivative equals the derivative of 4. Let's identify each derivative starting with X 2 Y. We're going to differentiate X2 X. Multiplied by the derivative of X because now x is a function of y. Multiplied by Y. And then according to the product rule, we are multiplying X2. By the derivative of y. And the derivative of Y is simply 1, so we get 2 X X Y plus X2. Now for our next derivative, we want to differentiate XY 2. Which gives us the derivative of X, that's X. Multiplied by Y2. Plus X. Multiplied by the derivative of y squared. And this is X Y2 plus. 2 xy because the derivative of Y2 is 22 in this case. So we have our derivatives, and now we can substitute everything in, so we have 2 multiplied by 2 X X Y. Which is 4 X X Y. Plus. 2 X2. Mindless 3 x prime. Y squared? Mindless We Multiplied by 2 XY, so that would be minus 6. X Y Equals 0. Once again, the derivative of 4 is 0. Now let's factor out X, which leaves us with 4 XY. Arancis. Minus 3y2d and we can move the other terms to the right so we get. 6 XY on the right. Minus 2 x 2. Solving for x we get. 6 XY. Minus 2 X squared. Divided by 4 XY. Minus 3 Y 2. And now what we can do is simply factor out the negative sign for any of our terms. For example, we can seek out the negative sign. For the numerator, which is going to give us negative in C2 X2 minus 6 XY, and the negative sign for the denominator. So, minus. 3 Y2 minus 4 XY. The reason why we're doing this is because now we can clearly see that those negative signs can be canceled out. We can remove the prey. And this allows us to visualize the relationship between the derivatives. This is D X divided by D Y. We can label our answer for this part, and now what we can notice is that these are reciprocals because now our original numerator has become the denominator, and the denominator is now the numerator and the derivative of E X divided by D Y. So our conclusion is that they are reciprocals. Let's label it and thank you for watching.

In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?

53. xy³ + x²y = 6

Key Concepts

Implicit Differentiation

Chain Rule

Reciprocal Relationship of Derivatives

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In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?53. xy³ + x²y = 6