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(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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00:39

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Campbell University

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

03:46

(a) Find $y^{\prime}$ by i…

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(a) Find $ y' $ by im…

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1-4(a) Find $y^{\prime…

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$1-4$(a) Find $y^{\pri…

06:11

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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