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Problem

(a) Find $ y' $ by implicit differentiation. (b)…

02:22

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Problem 3 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).

$ \sqrt{x} + \sqrt{y} = 1 $


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01:38

Frank Lin

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Related Topics

Derivatives

Differentiation

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Top Calculus 1 / AB Educators
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Derivatives - Intro

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.

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Differentiation Rules - Overview

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.

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Watch More Solved Questions in Chapter 3

Problem 1
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Problem 5
Problem 6
Problem 6
Problem 7
Problem 8
Problem 9
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Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
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Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 21
Problem 22
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Problem 24
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Problem 34
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Problem 76
Problem 77
Problem 78
Problem 79
Problem 80

Video Transcript

In this problem were given an equation, wording equation is of x, plus the root of y is equal to 1. In part, a we asked to use implicit differentiation to find very to function y. So let's stood out. Let'S take derivative of all the terms with respect to x. Preston would give us 1 over 2 square root of x. Plus second term would give us 1 over 2 square. Now we're in to inner function since y is also a function of have dx and that is equal to 0, because 1 is just a constant. So let's leave the i d x on the 1 side we have y t is then equal to negative 1. Over 2 star root of x, divided by 1 over 2 square root. I means to would cancel out and you would find the answer as the negative of y divided by s x now in part b, were asked to explicitly solve for y and then take derivative of that function. So we see that square root of y is equal to 1 minus square root of x. Now, if we take square of sides, we find y to be 1 minus square root of x. Squared now, let's find which of this y prime is equal to 2 times 1 minus square root of x times the roof inner function, which is negative 1 over 2 square root of x and 2, will cancel out and we find then the answer to be negative. So square root of x, minus 1 give i by square root of x all right now in part c, we are asked to compare the derivative we find in part and part b. So, let's take derivative of the function y that we find in part a so we have negative of the root of y divided by the square root of x. Now, in part, we find what our function y is, and from this we see that y is equal to negative of square root of 1 minus square root of x square square, root, divided by square x, square root and square will cancel out and will end up With negative of 1 minus square root of x, divided by square root of x, and that is equal to square root of x, minus 1 divided by square and as you can see, this is same as what we had for. So we can see that you can say that both implicit and explicit differentiation gi us the same result.

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Related Topics

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Video Thumbnail

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