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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 3.8.23b
Capitolo 3, Problema 3.8.23b

13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
³√x+³√y⁴ = 2;(1,1)

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1
Start by understanding the given equation: \( \sqrt[3]{x} + \sqrt[3]{y^4} = 2 \). This is an implicit function of \( x \) and \( y \).
Differentiate both sides of the equation with respect to \( x \). Remember to use the chain rule for \( \sqrt[3]{y^4} \), which involves differentiating \( y \) with respect to \( x \) (i.e., \( \frac{dy}{dx} \)).
The derivative of \( \sqrt[3]{x} \) with respect to \( x \) is \( \frac{1}{3}x^{-2/3} \). For \( \sqrt[3]{y^4} \), apply the chain rule: \( \frac{4}{3}y^{3} \cdot \frac{dy}{dx} \).
Set the derivative of the left side equal to the derivative of the right side, which is zero, since the derivative of a constant (2) is zero.
Substitute \( x = 1 \) and \( y = 1 \) into the differentiated equation to solve for \( \frac{dy}{dx} \), which represents the slope of the curve at the point (1,1).

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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. Instead of solving for one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for curves defined by equations that cannot be easily rearranged.
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Percorso guidato
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Finding The Implicit Derivative

Slope of a Curve

The slope of a curve at a given point represents the rate of change of the dependent variable with respect to the independent variable at that point. Mathematically, it is found by evaluating the derivative of the function at the specified coordinates. In the context of implicit differentiation, the slope can be determined by substituting the coordinates of the point into the derivative obtained from the implicit differentiation process.
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Summary of Curve Sketching

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that if a function y is defined as a function of u, which in turn is a function of x, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This rule is essential in implicit differentiation, where we often encounter nested functions.
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Intro to the Chain Rule
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Domanda del libro di testo

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>

b. Graph the tangent lines on the given graph.

x+y³−y=1; x=1

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Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

<IMAGE>

b. h′(2)h^{\(\prime\)}\(\left\)(2\(\right\))

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13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

(x+y)^2/3=y; (4, 4)

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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

b. d/dx(tan^−1 x) =sec² x

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Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = √3x; a= 12

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{Use of Tech} A mixing tank A 500-liter (L) tank is filled with pure water. At time t=0, a salt solution begins flowing into the tank at a rate of 5 L/min. At the same time, the (fully mixed) solution flows out of the tank at a rate of 5.5 L/min. The mass of salt in grams in the tank at any time t≥0 is given by M(t) = 250(1000−t)(1−10−³⁰(1000−t)¹⁰) and the volume of solution in the tank is given by V(t) = 500-0.5t.

b. Graph the volume function and verify that the tank is empty when t=1000 min. 

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