Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 3 - Derivatives

Problema 3.89

Capitolo 3, Problema 3.89

Find the slope of the curve x³y³ + y² = x + y at the points (1, 1) and (1, -1).

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To find the slope of the curve at a given point, we need to find the derivative of the curve with respect to x. The curve is given by the equation: x³y³ + y² = x + y.

Use implicit differentiation to differentiate both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, apply the chain rule.

Differentiate the left side: For x³y³, use the product rule: d/dx(x³y³) = x³ * d/dx(y³) + y³ * d/dx(x³). For y², use the chain rule: d/dx(y²) = 2y * dy/dx.

Differentiate the right side: d/dx(x) = 1 and d/dx(y) = dy/dx.

After differentiating, solve the resulting equation for dy/dx to find the expression for the slope of the curve. Substitute the points (1, 1) and (1, -1) into this expression to find the slope at each point.

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Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function defined implicitly by an equation involving both x and y. 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, which represents the slope of the curve at any point.

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Percorso guidato

05:14

Finding The Implicit Derivative

Slope of a Curve

The slope of a curve at a given point is defined as the rate of change of the y-coordinate with respect to the x-coordinate at that point. Mathematically, it is represented by the derivative of the function at that point. For curves defined by implicit equations, the slope can be found using the derivative obtained through implicit differentiation.

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11:41

Summary of Curve Sketching

Evaluating Derivatives at Specific Points

Once the derivative (dy/dx) is determined, evaluating it at specific points involves substituting the coordinates of those points into the derivative expression. This gives the slope of the curve at those specific locations. In this case, we will substitute the points (1, 1) and (1, -1) into the derived expression to find the respective slopes.

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04:50

Critical Points

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Using the Alternative Formula for Derivatives

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f'(x) = lim (z → x) (f(z) − f(x)) / (z − x)

to find the derivative of the functions in Exercises 23–26.

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Linearization for Approximation