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Find all points on the curve $ x^2 y^2 + xy = 2 $ where the slope of the tangent line is -1.

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00:52

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Missouri State University

University of Nottingham

Boston College

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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Find all points on the cur…

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Find the equation of the …

In this problem were given an equation of function and then we are asked to find all the points that the where the slope of tangent line is negative. 1. So, let's first find the slope of tangent line to do the elastic derivative of all the terms with respect to x. We have 2 x times y squared, so we can use product for this torment for the second term 2 x y square plus 2 x. Squared y times y prime plus y plus x times y is equal to 0 point. So from this we see that y prime times 2 x, squared y plus x, is equal to negative of 2 x squared plus y. It means that y prime will be of form negative y times 2 x y plus 1, divided by x times 2 x y plus 1. These will cancel out and we find the derivative to be negative y over x. We want look to be negative 1, so it means that negative y over x will be equal to negative 1 point from this. We see that y then should be equal to x. Now we're going to whenever we see why in the original equation, we're going to apply x and then we're going to solve for x. So let's do that. We have x, squared times x, squared plus x times x is equal to point. This is x square times x. Square plus 1 is equal to 2 point from this we see that x could be 1 or x could be negative. 1 point now using this when x is 1, we see that y is 1, and when i give to the point where the slope of tangent line is negative, 1 are 11 and negative or negative. 1 point.

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