58–59. Carry out the following steps.
a. Use implicit di...

Question

58–59. Carry out the following steps.

a. Use implicit differentiation to find dy/dx.

xy^5/2+x^3/2y=12; (4, 1)

Accepted Answer

Welcome back everyone. Given the equation X Y 7 plus X 5 Y equals 252, perform implicit differentiation to determine the derivative of Y with respect to X. No, because we're given the equation in this form, then it makes sense why we would use implicit differentiation. So to implicitly differentiate our equation, we can differentiate our left hand side and our right hand side with respect to X because we're finding the derivative of Y with respect to X, OK? So Let's just quickly write that here, so implicitly differentiating. OK, with respect to X. Starting with our left hand side, OK, notice that we have two terms here that are products. So we'll need to differentiate both of those using the product rule of differentiation. What does the product rule of differentiation say? Well, recall, it basically tells us that if we have a product Y equals UV, where U and V are functions of X, then the derivative of Y with respect to X. Eals U multiplied by V plus U multiplied by V, where U and V are the derivatives of U and V with respect to X. So let's apply that idea here to differentiate both of our terms, OK? Now, starting with XY to the part of 7 halves, OK. Then The derivative With respect to X of XY to the power of 7 houses, it's going to be equal to first U. If we say that U is X, then it's gonna be X multiplied by the derivative of Y to the power of 7 houses. So that's multiplied by Y. Sorry. Here we bring down our power 7 houses multiplied by Y to the power of 5 houses, DD X. OK. And now if we keep V Y to the power of 7 houses, then we can multiply it by the derivative of X which is 1, OK. On the other hand, If we differentiate the second term X to the power of 5 halves multiplied by Y, then we can keep you X to the power of 5 halves multiplied by the derivative of Y, OK, which is DYD X. Plus, the derivative of our second term, sorry, well, we keep our second term Y, OK, multiplied by the derivative of our first term X to the power of 5 halves, which is going to be 5 halves multiplied by X to the power of 3 halves, OK? So now we have the derivatives of both our terms, OK? And we know that the derivative of 252 with respect to X is going to be equal to zero. So no, that means by implicitly differentiating we should end up with 7 halves. X Y to the power of 5 halves D YD X. Plus Y to the power of 7 house for our first term and for the derivative of our second term, it's going to be X to the power of 5 house. DUIDX, OK. Multiplied by plus 5 house X to the power of 3 houses. Multiplied by y again for a second term and again we said the derivative of 252 with respect to x equals 0. Know that we have it this way to determine DDX we can try to solve for DUIDX from our expression. Now, first, let's group all of the terms that have DUIDX, OK? So, it's gonna be, let me highlight it here, it's gonna be our first term and our third term, OK? And then we're going to move our next two next terms to the other side of the equation. So now if we factor out the ID X, OK, we're going to be left. With X, well, let me add that here. 7 halves of X multiplied by 5 halves. Was to the power of Y? Well, it's why that's raised to the power of 5 has actually, so let me just write that properly here. The 7 halves of X to the of, sorry, XY to the power of 5 halves, OK. Plus our term here X to the power of 5 halves, OK, is going to be equal to, let's bring our next two terms to the other side. Of the equation, so that's gonna be equal to negative Y to the power of 7 minus 5/X to the power of 3/ Y, OK. And no. To solve for DUIDX, then we can divide both sides by this expression that's on the the right hand side, OK. Or the left hand side, sorry, so that means D X D X equals negative Y to the part of 7 halves. OK, minus 5 X X 3 Y. I'll do, I But 7 houses of XY. Yeah. I'm just writing this all clearly here. So 7 house of XY to the power of 5 houses, plus X to the power of 5 houses. Now, we could simplify this a bit further because notice that we have a denominator of two existing in our numerator and or a denominator for fractions, OK? So now if we multiply both the denominator and numerator by 2. Then our problem is gonna simplify to negative 2 Y 7/2 minus 5 X to the 3/al Y, OK, divided. By 7 xy to the power of 5 plus 2 X to the power of 5 halves. And now notice that we can factor Y from our numerator and X from our denominator respectively, OK? And when we do that in our denominator it's going to become, well, negative, OK. So we're gonna have negative factor Y. I'm gonna be left with 2 y to the power of 5 halves plus 5 X to the power of 3 halves and of course that's because we factored out our negative term here and in the denominator we'll have X that's being multiplied by 7 y to the 5 halves plus 2 X to the power of 3 halves. Thus this is going to be the derivative of Y with respect to X. Thanks a lot for watching everyone. I hope this video helped.

58–59. Carry out the following steps.a. Use implicit differentiation to find dy/dx.xy^5/2+x^3/2y=12; (4, 1)

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58–59. Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
xy^5/2+x^3/2y=12; (4, 1)