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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 3 - DerivativesProblema 3.8.48a
Capitolo 3, Problema 3.8.48a

45–50. Tangent lines Carry out the following steps. <IMAGE>
a. Verify that the given point lies on the curve.
x⁴-x²y+y⁴=1; (−1, 1)

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First, substitute the given point (-1, 1) into the equation of the curve x⁴ - x²y + y⁴ = 1 to verify if it satisfies the equation.
Calculate the left-hand side of the equation by substituting x = -1 and y = 1: (-1)⁴ - (-1)²(1) + (1)⁴.
Simplify the expression: 1 - 1 + 1.
Evaluate the simplified expression to check if it equals the right-hand side of the equation, which is 1.
If the left-hand side equals the right-hand side, then the point (-1, 1) lies on the curve. Otherwise, it does not.

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Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. In this case, the equation x⁴ - x²y + y⁴ = 1 involves both x and y, making it necessary to apply the chain rule when differentiating terms involving y. This method allows us to find the derivative dy/dx without solving for y explicitly.
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05:14
Finding The Implicit Derivative

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point and has the same slope as the curve at that point. The slope of the tangent line can be found using the derivative of the function at that point. For the curve defined by the equation, once we find dy/dx, we can evaluate it at the point (−1, 1) to determine the slope of the tangent line.
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Slopes of Tangent Lines

Point Verification

Verifying that a point lies on a curve involves substituting the coordinates of the point into the equation of the curve. If the left-hand side of the equation equals the right-hand side after substitution, the point is confirmed to be on the curve. In this case, substituting (−1, 1) into the equation x⁴ - x²y + y⁴ = 1 will confirm whether this point lies on the curve before proceeding with further calculations.
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Critical Points
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Find an equation of the line tangent to the following curves at the given value of x.

y = csc x; x = π/4

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13-26 Implicit differentiation Carry out the following steps.

a. Use implicit differentiation to find dy/dx.

x = e^y; (2, ln 2)

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45–50. Tangent lines Carry out the following steps. <IMAGE>

a. Verify that the given point lies on the curve.

(x²+y²)²=25/4 xy²; (1, 2)

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