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At what point on the curve $ y = \sqrt {1 + 2x} $ is the tangent line perpendicular to the line $ 6x + 2y = 1? $

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01:10

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Derivatives

Differentiation

Chanse S.

November 15, 2020

Thank you!! You teach at Oregon State?

Campbell University

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University of Michigan - Ann Arbor

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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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At what point on the curve…

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in this problem. We're looking for the point where the Tanja line is perpendicular to the given line and perpendicular lines have opposite reciprocal slopes. So let's find the slope of the given line. Let's take this equation and isolate why. So we'll subtract six x from both sides and divide by two. So the slope of this line is negative. Three. So the perpendicular slope would be positive 1/3. So we're looking for the point where the Tanja line has a slope of 1/3. That means we want the derivative to equal 1/3. So let's find the derivative. I'm going to rewrite the function as one plus two x to the 1/2 power and then I'm going to use the chain rule. Why Prime equals 1/2 times one plus two x to the negative 1/2 power that takes care of the first part of the chain rule times the derivative of the inside, which is to now we can multiply the 1/2 of the two that was canceled. And now we want this derivative to be equal to 1/3 so this derivative could be written as one over the square root of one plus two x, and it equals 1/3. Let's take the reciprocal of both sides, and we have a square root of one plus two. X is equal to three. Now it's square both sides, and we have one plus two. X equals nine. Subtract 12 X equals eight and divide by two. X equals four. Okay, this is the X coordinate of the point. We also want the Y coordinate of the point, and the Y coordinate would be what we get when we take four and substituted into the original equation. So we have one plus two times for inside a square root square root of nine is three. So the point we're looking for is the 0.0.43

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