Domanda del libro di testo
Find all points (x, y) on the graph of y = x/(x − 2) with tangent lines perpendicular to the line y = 2x + 3.
241
views
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 3, Problema 44
Find by implicit differentiation.
x² + xy + y² - 5x = 2
Guida verificata passo dopo passo1
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 term x² with respect to x, which gives 2x.
For the term xy, apply the product rule: differentiate x to get 1, multiply by y, then differentiate y with respect to x (dy/dx) and multiply by x. This results in y + x(dy/dx).
Differentiate y² with respect to x using the chain rule: 2y(dy/dx).
Differentiate the term -5x with respect to x, which gives -5. Set the derivative of the right side, which is a constant 2, to 0. Combine all these results to form the equation: 2x + y + x(dy/dx) + 2y(dy/dx) - 5 = 0.
Risposta video verificata per un problema simile:
Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
5mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent variable is not isolated on one side. Instead of solving for y explicitly, 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 needing to express y in terms of x.
Video consigliato:
Percorso guidato
05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When applying implicit differentiation, we often encounter terms involving y, which requires us to use the chain rule to differentiate these terms correctly. For example, when differentiating y², we apply the chain rule to get 2y(dy/dx).
Video consigliato:
05:02Intro to the Chain Rule
Collecting Like Terms
Collecting like terms is a crucial step in simplifying equations after differentiation. After applying implicit differentiation, we often end up with an equation that contains multiple terms involving dy/dx. By rearranging and combining these terms, we can isolate dy/dx, making it easier to solve for the derivative and understand the relationship between x and y.
Video consigliato:
4:46Solving Exponential Equations Using Like Bases
Pratica correlata
Domanda del libro di testo
Slopes and Tangent Lines
a. Horizontal tangent lines Find equations for the horizontal tangent lines to the curve y = x³ − 3x − 2. Also find equations for the lines that are perpendicular to these tangent lines at the points of tangency.
329
views
Domanda del libro di testo
Find by implicit differentiation.
x²y² = 1
433
views
Domanda del libro di testo
Quadratics having a common tangent line The curves y = x² + ax + b and y = cx − x² have a common tangent line at the point (1,0). Find a, b, and c.
251
views
Domanda del libro di testo
Slopes and Tangent Lines
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope?
298
views
Domanda del libro di testo
Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10/9)t², where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?
220
views