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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 3 - DerivativesProblema 3.8.26a
Capitolo 3, Problema 3.8.26a

13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
(x+y)^2/3=y; (4, 4)

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Start by differentiating both sides of the equation with respect to x. The equation given is \((x + y)^{2/3} = y\). Apply the chain rule to differentiate \((x + y)^{2/3}\) with respect to x.
The chain rule states that if you have a composite function \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\). Here, let \(u = x + y\), so \((x + y)^{2/3}\) becomes \(u^{2/3}\). Differentiate \(u^{2/3}\) with respect to u, which gives \(\frac{2}{3}u^{-1/3}\).
Now, differentiate \(u = x + y\) with respect to x. This gives \(\frac{du}{dx} = 1 + \frac{dy}{dx}\). Substitute back to get the derivative of \((x + y)^{2/3}\) with respect to x: \(\frac{2}{3}(x + y)^{-1/3}(1 + \frac{dy}{dx})\).
Differentiate the right side of the equation \(y\) with respect to x, which is simply \(\frac{dy}{dx}\).
Set the derivatives equal: \(\frac{2}{3}(x + y)^{-1/3}(1 + \frac{dy}{dx}) = \frac{dy}{dx}\). Solve this equation for \(\frac{dy}{dx}\) to find the implicit derivative.

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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 y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This method is particularly useful for equations that are difficult or impossible to rearrange.
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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 when differentiating terms involving y, as we must multiply by dy/dx to account for the dependence of y on x. This is crucial for correctly finding the derivative of expressions where y is not isolated.
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Intro to the Chain Rule

Evaluating Derivatives at a Point

After finding the derivative dy/dx using implicit differentiation, we often need to evaluate it at a specific point, such as (4, 4) in this case. This involves substituting the x and y values into the derived expression to find the slope of the tangent line at that point. This step is essential for understanding the behavior of the function at specific coordinates.
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Critical Points
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