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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 3 - DerivativesProblema 3.8.50b
Capitolo 3, Problema 3.8.50b

45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
(x²+y²)²=25/4 xy²; (1, 2)

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First, identify the given implicit equation of the curve: \((x^2 + y^2)^2 = \frac{25}{4} xy^2\). We need to find the derivative \(\frac{dy}{dx}\) using implicit differentiation.
Differentiate both sides of the equation with respect to \(x\). For the left side, use the chain rule: \(2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\). For the right side, apply the product rule: \(\frac{25}{4} (y^2 + 2xy \frac{dy}{dx})\).
Set the derivatives equal: \(2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = \frac{25}{4} (y^2 + 2xy \frac{dy}{dx})\).
Substitute the given point \((1, 2)\) into the differentiated equation to solve for \(\frac{dy}{dx}\). This will give the slope of the tangent line at the point.
Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step and \((x_1, y_1) = (1, 2)\), to write the equation of the tangent line.

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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 (x² + y²)² = (25/4)xy² involves both x and y, requiring us to differentiate both sides with respect to x while treating y as a function of x. This method allows us to find dy/dx, which is essential for determining the slope of the tangent line.
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Percorso guidato
05:14
Finding The Implicit Derivative

Tangent Line Equation

The equation of a tangent line at a given point on a curve can be expressed using the point-slope form: y - y₀ = m(x - x₀), where (x₀, y₀) is the point of tangency and m is the slope at that point. Once the derivative (slope) is calculated using implicit differentiation, it can be substituted into this formula along with the coordinates of the point (1, 2) to find the specific equation of the tangent line.
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Percorso guidato
05:14
Equations of Tangent Lines

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. In the context of implicit differentiation, it allows us to differentiate terms involving y, which is a function of x. For example, when differentiating y², we apply the chain rule to obtain 2y(dy/dx), which is crucial for correctly finding the derivative of the given implicit equation.
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Intro to the Chain Rule
Pratica correlata
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Derivatives and tangent lines

a. For the following functions and values of a, find f′(a).

f(x) = 8x; a = −3

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Derivatives of inverse functions from a table Use the following tables to determine the indicated derivatives or state that the derivative cannot be determined. <IMAGE>

b. (f^-1)'(6)

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Use a graphing utility to graph the curve and the tangent line on the same set of axes.

y = (x + 5) / (x - 1); a = 3

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79–82. {Use of Tech} Visualizing tangent and normal lines

b. Graph the tangent and normal lines on the given graph.

(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

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{Use of Tech} Computing limits with angles in degrees Suppose your graphing calculator has two functions, one called sin x, which calculates the sine of x when x is in radians, and the other called s(x), which calculates the sine of x when x is in degrees.

b. Evaluate lim x→0 s(x) / x. Verify your answer by estimating the limit on your calculator.

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21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(t) = 3t⁴; a= -2, 2

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