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Find $ dy/dx $ by implicit differentiation. $ …

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Problem 4 Medium Difficulty

(a) Find $ y' $ by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get $ y' $ in terms of $ x. $
(c) Check that your solutions to part (a) and (b) are consistent by substituting the expression for $ y $ into your solution for part (a).

$ \frac {2}{x} - \frac {1}{y} = 4 $

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Video Transcript

In this trouble were given an equation or equation is 2 x, minus 1 or y is equal to 4. In part, a we are asked to use implicit differentiation to find derivative of y. So let's take derivative of whole terms with respect to x. The first term would give us negative 2. Squared second term would give us plus 1 y squared, don't forget. Y is also a functional x or we have divined the x term here, and that is 0, since 4 is just the question. But from this we see that then di dx, so there were to appoint with respect to x, is equal to 2 over x squared times y squared. So that is 2 y squared divided by x, squared in part, were asked to find the function y and then take care of that. We see that 1 over y is equal to 2 over x minus 4 point from this. We see that the y is equal to 2 over x minus 4 inverell right now. Let'S take the rivatthat y prime, isn't equal to negative 1 times 2 over x minus 4 to the negative second power times the wittier function, which is negative 2 over x squared. So the answer is then 2 over x, squared multiplied by 2 over x minus 4 to the second negative second power all right now, in part c, we are asked to compare the derivatives we find in part a and b. So, let's take derivative that we find in part a we have 2 y squared divided by x square. Now, in part b, we find what a is right. This is our function as why don't we just plug that in? If we do so, we see that this is 2 over x square multiplied by y square, where y squared is 2 over x, minus 4 negative 1 point. This is y squared of this, and from this we see that derivative should be to xpres multiplied by 2. Over x, minus 4 to the negative second power and y can see it's the same as what they found in part b. So it means that implicit and explicit derivatives or differentiation, give us the same results.

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Find $ dy/dx $ by implicit differentiation. $ …

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James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals