Textbook Question
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
f(x) = √(x + 1), (8, 3)
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.7.1
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x²y + xy² = 6
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, use implicit differentiation.
Differentiate the first term x²y. Apply the product rule: d/dx(x²y) = x²(dy/dx) + y(2x).
Differentiate the second term xy². Again, use the product rule: d/dx(xy²) = x(2y(dy/dx)) + y².
Differentiate the right side of the equation, which is a constant: d/dx(6) = 0.
Combine all the differentiated terms: x²(dy/dx) + y(2x) + x(2y(dy/dx)) + y² = 0. Solve for dy/dx by isolating it on one side of the equation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This allows us to find dy/dx without isolating y, which is particularly useful for complex relationships.
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05:14
Finding The Implicit Derivative
Product Rule
The product rule is a fundamental differentiation rule used when differentiating the product of two functions. It states that if u and v are functions of x, then the derivative of their product is given by d(uv)/dx = u'v + uv'. In the context of implicit differentiation, this rule is essential when differentiating terms that involve products of x and y, ensuring that both functions are accounted for.
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05:18The Product Rule
Chain Rule
The chain rule is a key principle in calculus that allows us to differentiate composite functions. It states that if a function y is dependent on u, which in turn is dependent on x, then dy/dx = (dy/du) * (du/dx). In implicit differentiation, the chain rule is applied when differentiating terms involving y, as we must multiply by dy/dx to account for the dependence of y on x.
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05:02Intro to the Chain Rule
Related Practice
Textbook Question
Tangent line to y = √x Does any tangent line to the curve y = √x cross the x-axis at x = −1? If so, find an equation for the line and the point of tangency. If not, why not?
Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x³ + y³ = 18xy
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
q = sin(t / (√t + 1))
Textbook Question
Derivative Calculations
In Exercises 1–8, given y = f(u) and u = g(x), find dy/dx = f'(g(x)) g'(x).
y = sin u, u = 3x + 1
Textbook Question
Find the points on the curve y = tan x, -π/2 < x < π/2, where the normal line is parallel to the line y = -x/2. Sketch the curve and normal lines together, labeling each with its equation.