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?

54. x³ + y² = sin²y

Accepted Answer

Welcome back, everyone. Given the equation X2 + Y cubed equals cosine quad X, find both Dy divided by DX and DX divided by D Y and describe how these derivatives relate to each other. Let's begin by finding the Y divided by D X. And this means that we're assuming that Y is a function of X. Let's use implicit differentiation starting with the derivative of X2. We're going to add the derivative of Y cubed. And this must be equal to the derivative of the right hand side, or the derivative of cosine squared of X. Now, X is the independent variable, so the derivative of X2 is going to be 2 X based on the power rule. The derivative of Y cubed is 3 Y2d based on the power rule, followed by Y based on the chain rule because Y is a function of X. On the right hand side, we have a composite function. We have to begin with the power rule, which gives us two cosine of X. Followed by the chain rule, we have to multiply by the derivative of cosine which is negative sine. Of X Now, if we simplify, we're going to get 2 X plus 3 Y 2 Y. Equals -2 x x. Cosine x. And negative to sin x cosine x, according to the double angle formula is simply. Negative sign of two acts. What we can do from here is solve for Y, so let's rearrange the expression. We get 3 Y 2 Y. Equals Now we have negative sign. Of 2 X and we have to subtract 2 X from both sides, right? So solving for Y we can show that Y equals negative. Let's factor out the negative sign. Of 2 Xs. Now we're going to say plus because we already have factored out the negative sign. Sign of two acts. Divided by 3 Y squad. And that's our the Y divided by the X. So let's label our first answer to this problem. Now let's move on to the second part. Where we are going to identify DX divided by D Y. So now X is a function of Y. We're going to perform the same first step. Simply let's write down the derivatives that we once evaluate. And now let's begin differentiating so the derivative of X is going to be two Xs based on the power rule, but now X is a function of Y. So we're also multiplying by the derivative of X because Well, we need to apply the chain rule for a composite function, because in this case, X is not an independent variable anymore. But Y is the independent variable in this case, which means that the derivative of Y cubed is simply 3y2d. So we're not multiplying by Y here because Y is the independent variable. For the derivative of cosine squared of X we're going to get the same two cosine of x. Multiplied by a negative sign of X. But then, since X is a function of Y, we also need to multiply by X, right, which is The derivative of our function X since X is a function of Y, we need to apply the chain rule. And that we can once again solve for the derivative, and in this case that would be X. First of all, we're going to show that. 2 X X plus. We Y2 is going to be equal to negatives. Of 2 X multiplied by X, so we can also write X. In front And now let's combine the X terms on the same side so we can show that 3 y squad. Is equal to negative X. Sign of 2 X minus 2 XX. If we perform factorization, we can show that this is negative X and Cs. Sign of 2 X plus 2 X. And now solving for X we can show that X is equal to negative. We're going to have a fraction. 3 y squad. Divided by 2 acts Plus sign of 2 X. And that's DX divided by D Y. So we have our second answer, and now we can notice that these are reciprocals because for our 2nd derivative. The original denominator is now our numerator. And the numerator has become the denominator, so we can also conclude that they are reciprocals. And that's the 3rd answer to this problem. 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?

54. x³ + y² = sin²y

Key Concepts

Implicit Differentiation

Inverse Function Theorem

Chain Rule

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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?54. x³ + y² = sin²y