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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.43

Implicit Differentiation


In Exercises 43–50, find by implicit differentiation.


xy + 2x + 3y = 1

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 term xy. Use the product rule: if u = x and v = y, then the derivative of uv is u'v + uv'. Here, u' = 1 and v' = dy/dx.
Differentiate the term 2x. The derivative of 2x with respect to x is simply 2.
Differentiate the term 3y. Since y is a function of x, use the chain rule: the derivative is 3(dy/dx).
Combine all the differentiated terms and set them equal to the derivative of the constant on the right side of the equation, which is 0. Solve for dy/dx to find the derivative of y with respect to x.

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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 to account for the dependent variable's implicit relationship.
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Finding The Implicit Derivative

Chain Rule

The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When using implicit differentiation, the chain rule is applied to terms involving the dependent variable, treating it as a function of the independent variable. This means that when differentiating a term like y, we multiply by dy/dx, the derivative of y with respect to x.
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Intro to the Chain Rule

Solving for dy/dx

After applying implicit differentiation to an equation, the next step is to isolate dy/dx to find the derivative of y with respect to x. This involves rearranging the differentiated equation to express dy/dx in terms of x and y. This process is crucial for understanding how y changes in relation to x, especially in contexts where y cannot be easily expressed as a function of x.
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Solving Logarithmic Equations