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If $ xy + e^y = e, $ find the value of $ y" $ at the point where $ x = 0. $

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02:16

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Harvey Mudd College

Baylor 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.

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In this problem were given an equation and we're justified secondary to apply with respect to x and x is 0 point. So, let's first plug x, 0 in and let's see the value of the function y when x is 0. So we have 0 times y plus e to the y is equal to e. This term is 0. So from this we see that when x is 0 y is equal to 1. All right now to find the first zertho have less differentiate. Both sides of disrectxwe have y plus x times y prime plus e to the y times y prime is equal to 0 point from this. We see that then y prime is equal to negative y divided by x, plus e to y. When x is 0, we found that y is equal to 1 plus evoid derivative. At that point, so y prime would then be equal to negative 1 over 0 plus e to the 1, and that will be negative 1 over e all right now, using this y prime equation, we're going to differentiate the 1 more time to find second dative. We only use question rule o y double prime would then be equal to the negative of y prime times x, plus e to the y plus y times 1 plus e to the y times y prime divided by x, plus e to the y square. All right now, let's simulate y double prime secondary to y 1 x 0, that is negative, vita y prime, is equal to negative 1, or so we have negative 1 over e times 0 plus e plus 1 times 1 plus e times negative 1 over e divided By 0 plus e square now these will cancel out. This will cancel out and we don't find y double prime to be negative of negative 1 over square. So the answer is 1 over b square.

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