Skip to main content
Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 3 - DerivativesProblema 3.7.12
Capitolo 3, Problema 3.7.12

Differentiating Implicitly


Use implicit differentiation to find dy/dx in Exercises 1–14.


x⁴ + sin y = x³y²

Guida verificata passo dopo passo
1
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 the chain rule.
Differentiate the left side: The derivative of x⁴ with respect to x is 4x³. For sin(y), use the chain rule: the derivative is cos(y) * dy/dx.
Differentiate the right side: Use the product rule for x³y². The product rule states that d(uv)/dx = u'v + uv'. Here, u = x³ and v = y². Differentiate u to get 3x² and v to get 2y * dy/dx.
Substitute the derivatives back into the equation: 4x³ + cos(y) * dy/dx = 3x²y² + x³ * 2y * dy/dx.
Rearrange the equation to solve for dy/dx. Collect all terms involving dy/dx on one side and factor dy/dx out. Then, solve for dy/dx by isolating it on one side of the equation.

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:
6m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. It involves differentiating both sides of an equation with respect to a variable, often x, while treating other variables, like y, as implicit functions of x. This method is essential when dealing with equations where y cannot be easily isolated.
Video consigliato:
Percorso guidato
05:14
Finding The Implicit Derivative

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In implicit differentiation, the chain rule is crucial when differentiating terms involving y, as it accounts for the derivative of y with respect to x.
Video consigliato:
05:02
Intro to the Chain Rule

Trigonometric Derivatives

Trigonometric derivatives are the derivatives of trigonometric functions, such as sine, cosine, and tangent. For example, the derivative of sin(y) with respect to y is cos(y). When using implicit differentiation, it's important to apply these derivatives correctly, especially when differentiating terms like sin(y) with respect to x, which involves using the chain rule to account for dy/dx.
Video consigliato:
06:35
Derivatives of Other Inverse Trigonometric Functions
Pratica correlata
Domanda del libro di testo

In Exercises 41–58, find dy/dt.


y = √(3t + (√2 + √(1 − t)))

288
views
Domanda del libro di testo

Finding Derivative Functions and Values 


Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.


g(t) = 1/t²; g′(−1), g′(2), g′(√3)

227
views
Domanda del libro di testo

Find the derivatives of the functions in Exercises 17–28.


y = ((x + 1)(x + 2)) / ((x − 1)(x − 2))

191
views
Domanda del libro di testo

Assume that functions f and g are differentiable with f(2) = 3, f'(2) = −1, g(2) = −4, and g'(2) = 1. Find an equation of the line perpendicular to the line tangent to the graph of F(x) = (f(x) + 3) / (x − g(x)) at x = 2.

220
views
Domanda del libro di testo

Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.


x ƒ(x) g(x) ƒ'(x) g'(x)

0 1 1 -3 1/2

1 3 5 1/2 -4


Find the first derivatives of the following combinations at the given value of x.


a. 6ƒ(x) - g(x), x = 1

224
views
Domanda del libro di testo

Find the derivatives of the functions in Exercises 19–40.


p = √(3 − t)

264
views