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Use implicit differentiation to find an equation of the tangent line to the curve at the given point.$ x^{\frac {2}{3}} + y^{\frac {2}{3}} = (-3 \sqrt{3}, 1), $ (astroid)
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01:20
Frank Lin
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 5
Implicit Differentiation
Derivatives
Differentiation
Campbell University
Oregon State University
University of Nottingham
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, we are asked to use implicit differentiation to find the equation of tangent line to the given curve at just given point negative 3 state, 3 and 1. We know that equation of tangent line will be able to form y minus 1 is equal to the derivative of this function. Alice at given point multiply by x, minus x 9. So, in order to find the equation that we first need to find derivative lent at this given point all right: let's take derivative of this function with respects x. First, we have 2 third x, negative 1, third plus 2 or 3 y negative 1 third times y prime is equal to 0. So from this we see that y prime is the negative x over y to the power negative 1. Third point: since we know what x notariat is, we can calculate derivative at this given point so then that will be the derivative it will at that, given point will then be negative of x is 3 square 3, so that is negative 3 square root of 3 Times negative to 1. Third, that is equal to square root of 3, cube to the power negative 1 third, and that is equal to 1 over 33. Then, since now we know derivative at a given point, we can find the question of this line. That will be y. Minus 1 is equal to 1 over square root of 3 x, plus 3 square root of 3, and from this we dont fine equation to be square root of x. Sorry, we find the equation to be 1 over square root of 3 x. 1, over square root of b x, plus 4 pi.
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