Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.8.58a

Chapter 3, Problem 3.8.58a

58–59. Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
xy^5/2+x^3/2y=12; (4, 1)

Verified step by step guidance

Start by differentiating both sides of the equation with respect to x. Remember that y is a function of x, so you'll need to use implicit differentiation. The equation is: \( xy^{\frac{5}{2}} + x^{\frac{3}{2}}y = 12 \).

Apply the product rule to the term \( xy^{\frac{5}{2}} \). The product rule states that \( \frac{d}{dx}[u \cdot v] = u'v + uv' \). Here, let \( u = x \) and \( v = y^{\frac{5}{2}} \). Differentiate each part: \( u' = 1 \) and \( v' = \frac{5}{2}y^{\frac{3}{2}} \cdot \frac{dy}{dx} \).

Differentiate the second term \( x^{\frac{3}{2}}y \) using the product rule again. Let \( u = x^{\frac{3}{2}} \) and \( v = y \). Differentiate each part: \( u' = \frac{3}{2}x^{\frac{1}{2}} \) and \( v' = \frac{dy}{dx} \).

Combine the results from the differentiation steps. You will have an equation involving \( \frac{dy}{dx} \). Collect all terms involving \( \frac{dy}{dx} \) on one side of the equation and factor out \( \frac{dy}{dx} \).

Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. Substitute the point (4, 1) into the equation to find the specific value of \( \frac{dy}{dx} \) at that point.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not isolated on one side. Instead of solving for y explicitly, we differentiate both sides of the equation with respect to x, applying the chain rule when necessary. This method is particularly useful for equations that are difficult or impossible to rearrange.

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 defined as a function of u, which in turn is a function of x, 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 is essential in implicit differentiation when dealing with terms involving y.

Evaluating Derivatives at a Point

After finding the derivative dy/dx using implicit differentiation, it is often necessary to evaluate this derivative at a specific point, such as (4, 1) 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 crucial for understanding the behavior of the function at specific coordinates.

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Textbook Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

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Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.

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Textbook Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

f(t) = 3t⁴; a= -2, 2

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Use the graph of f in the figure to do the following. <IMAGE>

a. Find the values of x in (-2,2) at which f is not continuous.