Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problema 3.8.8

Capitolo 3, Problema 3.8.8

If x²y³ = 4/27 and dy/dt = ¹/₂, then what is dx/dt when x = 2?

Guida verificata passo dopo passo

First, recognize that the given equation x²y³ = 4/27 is an implicit function of x and y. We need to differentiate both sides of this equation with respect to time t to find the relationship between dx/dt and dy/dt.

Apply implicit differentiation to the equation x²y³ = 4/27 with respect to t. Use the product rule and chain rule: d/dt(x²y³) = d/dt(4/27). Since 4/27 is a constant, its derivative is 0.

Differentiate the left side: d/dt(x²y³) = 2x(dx/dt)y³ + x²(3y²)(dy/dt). This uses the product rule: d(uv)/dt = u(dv/dt) + v(du/dt), where u = x² and v = y³.

Substitute the known values into the differentiated equation: x = 2, dy/dt = 1/2, and solve for dx/dt. The equation becomes: 2(2)(dx/dt)y³ + (2)²(3y²)(1/2) = 0.

Solve the resulting equation for dx/dt. Rearrange the terms to isolate dx/dt and substitute any additional known values to find the expression for dx/dt.

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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. In this problem, the equation x²y³ = 4/27 involves both x and y, so we differentiate both sides with respect to time t, applying the chain rule to account for the derivatives of x and y.

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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 times the derivative of the inner function. In this context, it helps differentiate terms like x² and y³ with respect to t, considering dx/dt and dy/dt.

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05:02

Intro to the Chain Rule

Solving for a Derivative

Once the implicit differentiation is performed, the next step is to solve for the desired derivative, dx/dt. This involves substituting known values, such as x = 2 and dy/dt = 1/2, into the differentiated equation and isolating dx/dt. This process requires algebraic manipulation to express dx/dt in terms of the given quantities.

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5:02

Solving Logarithmic Equations

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