In Exercises 53 and 54, find dr/ds.

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Question

In Exercises 53 and 54, find dr/ds.

2rs - r - s + s² = -3

Accepted Answer

Welcome back everyone. Find DM divided by DN by implicit differentiation. We're given 2 mn plus 4 M minus 3N + n2d. We have to understand that in this problem, M is a function of N. Right, and what we're going to do is simply look at the given equation and differentiate both sides. We're going to take the derivative of the left hand side to mm + 4 m minus 3M + m quad. We're going to differentiate it and set it equal to the derivative of the right hand side, so the derivative of 7. 1st of all, we have to understand that we're differentiating the product mn because we can factor out the constant. Where adding 4 multiplied by the derivative of M, subtracting 3 multiplied by the derivative of N, and then adding the derivative of N2, which is going to be equal to the derivative of 7. So let's evaluate the derivative of mm. According to the product rule, we have to take the derivative of the first term, so that's our function m, which is then multiplied by the second function M, and then we are adding the first function, which is M multiplied by the second function's derivative. So N, right? And here we have applied the product rule. So M remains as it is because that's our DM divided by DN. We can simply write MN plus, and now the derivative of N is equal to 1, according to the power rule. That's how we're in the dependent variable. So we're simply adding M. And now going back into our original expression, this means that we get 2 multiplied by MN plus M. Plus 4 M. Minus 3 because the derivative of n is equal to 1 and the derivative of n squared is 2n according to the power rule. Now the derivative of 7 is equal to 0. From here, our goal is to solve for M because M is DM divided by DN. It's just a different notation, but we're looking for the derivative of M with respect to M. What we're going to do is simply distribute the terms so that 2MN + 2 M + 4 M minus 3 + 2 N equals 0. And now we're going to back through M. so we get M and 2 M. Plus 4, we're done, right, because we only had two terms with M. For all of the remaining terms, we basically want to move them to the right hand side of the equation. So first of all, we can move -3 to the right, which gives us positive 3. Then we can move 2n to the right, which gives us -2M, and we can move 2M to the right which gives us negative 2 M. And now we are simply solving for M, which means that we're dividing the right hand side, we. -2 minus 2 M. We're dividing that term by the factor adjacent to M, right? So that factor is 2 M + 4. Well then, this is our derivative of M with respect to M. We can label it as DM divided by DN. Let's also label our final answer in a box, and thank you for watching.

In Exercises 53 and 54, find dr/ds.

2rs - r - s + s² = -3

Key Concepts

Implicit Differentiation

Chain Rule

Solving for Derivatives

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