In Exercises 43–50, find by implicit differentiation.
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Question

In Exercises 43–50, find by implicit differentiation.

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√xy = 1

Accepted Answer

Welcome back, everyone. By using implicit differentiation, find D Y divided by D X for the curve given by the equation square root of XY equals x minus y. If we want to apply implicit differentiation, we basically want to differentiate the left hand side and the right hand sides. We're setting the derivative of the left hand side equals the derivative of the right hand side. Let's begin with the derivative of the left hand side. We can rewrite square root of X Y S Xy raises the power of 1/2 and apply the power rule, which says that first of all we bring down negative, we bring down positive 12, multiply that by xy, raise the power of negative 1/2 because we are subtracting one from the original exponent. And now because we have an inner function, we have to apply the chain rule and multiply the result by the derivative of X Y. So now xy is a product, so we need to apply the product rule. The product rule says that we're taking the derivative of X multiplying by Y and adding X multiplied by the derivative of Y. And now our function is Y of X, meaning X is the independent variable, so the derivative of X is 1, and we can simplify the resultant derivative as Y plus xy derivative. So we can now see that the derivative of the left hand side is 1/2 multiplied by xy raises the power of -1/2 multiplied by Y plus XY derivative, right? And this must be equal to the derivative of X minus Y. The derivative of X is 1, and the derivative of Y can be written as Y. So we're subtracting Y. From here we're going to perform simplification so we can say that first of all we have 1 divided by 2 square root of xy. That's because we have a negative exponent. This means that we can use the reciprocal rule and rewrite xy to the power of 1/2 in the denominator and because 1.5% square root, we're just adding square root and combining it with 12. We are multiplying that by Y + xy so we can write those in the numerator. So let's go ahead and write Y + xy in the numerator, and this is equal to 1 minus Y. From here we simply want to solve or y and notice that we have Y on both sides of the equation. So what we're going to do is simply combine the Y terms on the same side of the equation. And now how can we do that? Well, We can simply multiply both sides by the denominator assuming it is not equal to 0. So we're going to say that 1 minus Y. Multiplied by 2. Square root of xy so we can put 2 in front followed by square root of xy. This must be equal toy plus xy assuming that square root of xy is not equal to 0. So now let's distribute the terms we're going to have 2. square root of x y minus 2 square root of X Y is equal to Y plus xy. Now let's combine the terms involving Y on the same side so we can just move our second term to the right and we're going to move Y to the left, right? So we are making sure that we're isolating Y. So now, if we consider the left hand side, on the left hand side we're going to have the original 2 square root of XY. We are now moving Y to the left. This means that it becomes negative, so we are subtracting Y, and we're setting this equal to XY. Plus 2 Y. Square root of x y. So what we want to do is simply factor out Y because we're solving for Y. And since we are solving for Y prime. We want to write the remaining factors, right? So the remaining factor will be X. Plus 2 square root of XY. So we have successfully factored out Y and then we're setting it equal to the left hand side, which is 2 square root of XY minus Y. Solving for Y, we have to divide both sides by the factor adjacent toy. So we're taking the right hand side to square root of XY. Minus Y, and we are dividing it by X + 2 square root of XY. So 2 square root of XY plus X goes into the denominator. And we know that Y represents the Y divided by DX. It's just a different notation. So we have arrived at the final answer to this problem. Thank you for watching.

In Exercises 43–50, find by implicit differentiation.
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√xy = 1

Key Concepts

Implicit Differentiation

Chain Rule

Product Rule

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