Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo

Briggs 3rd Edition

Ch. 3 - Derivatives

Problema 3.8.20a

Capitolo 3, Problema 3.8.20a

13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
tan xy = x+y; (0,0)

Guida verificata passo dopo passo

Start by differentiating both sides of the equation with respect to x. The equation is \( \tan(xy) = x + y \).

Apply the chain rule to differentiate \( \tan(xy) \). The derivative of \( \tan(u) \) with respect to u is \( \sec^2(u) \), so the derivative of \( \tan(xy) \) with respect to x is \( \sec^2(xy) \cdot (y + x \frac{dy}{dx}) \).

Differentiate the right side of the equation. The derivative of \( x \) with respect to x is 1, and the derivative of \( y \) with respect to x is \( \frac{dy}{dx} \).

Set the derivatives equal to each other: \( \sec^2(xy) \cdot (y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx} \).

Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. Substitute \( x = 0 \) and \( y = 0 \) into the equation to find the specific value of \( \frac{dy}{dx} \) at the point (0,0).

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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 as necessary. This method is particularly useful for equations involving multiple variables or when isolating one variable is complex.

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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. It states that if a function y is composed of another function u, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This rule is essential in implicit differentiation, as it helps to differentiate terms involving products of variables.

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Intro to the Chain Rule

Evaluating at a Point

Evaluating a derivative at a specific point involves substituting the coordinates of that point into the derivative expression. In the context of the given problem, after finding dy/dx using implicit differentiation, we substitute the point (0,0) to determine the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific values and for applications such as finding tangent lines.

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Critical Points

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