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Chapter 3: Problem 22

In Exercises \(17-28,\) find \(d y\) $$ x y^{2}-4 x^{3 / 2}-y=0 $$

Short Answer

Expert verified
\( dy = \left( \frac{6x^{1/2} - y^2}{2xy - 1} \right) dx \).

Step by step solution

01

Implicit Differentiation Preparation

Given the equation \( x y^2 - 4x^{3/2} - y = 0 \), differentiate both sides with respect to \( x \) implicitly. Remember that \( y \) is a function of \( x \), so use the chain rule where necessary.
02

Differentiate the First Term

Differentiate \( x y^2 \) using the product rule: \( \frac{d}{dx}(x y^2) = y^2 \frac{d}{dx}(x) + x \frac{d}{dx}(y^2) = y^2 + 2xy \frac{dy}{dx} \).
03

Differentiate the Second Term

Differentiate \( -4x^{3/2} \) with respect to \( x \): \( \frac{d}{dx}(-4x^{3/2}) = -4 \cdot \frac{3}{2}x^{1/2} = -6x^{1/2} \).
04

Differentiate the Third Term

Differentiate \( -y \) with respect to \( x \): \( \frac{d}{dx}(-y) = -\frac{dy}{dx} \).
05

Write the Resulting Equation

Combine all differentiated components into the equation: \[ y^2 + 2xy \frac{dy}{dx} - 6x^{1/2} - \frac{dy}{dx} = 0 \].
06

Solve for \( \frac{dy}{dx} \)

Rearrange the equation to solve for \( \frac{dy}{dx} \): \[ 2xy \frac{dy}{dx} - \frac{dy}{dx} = 6x^{1/2} - y^2 \]. This simplifies to \[ (2xy - 1) \frac{dy}{dx} = 6x^{1/2} - y^2 \].
07

Explicitly Solve for \( \frac{dy}{dx} \)

Divide both sides by \( (2xy - 1) \) to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{6x^{1/2} - y^2}{2xy - 1} \].
08

Calculate \( dy \)

Since \( dy = \frac{dy}{dx} \cdot dx \), substitute the expression for \( \frac{dy}{dx} \): \( dy = \left( \frac{6x^{1/2} - y^2}{2xy - 1} \right) dx \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
Understanding derivatives is key to mastering calculus. A derivative measures the rate at which a function changes at any point. In simpler terms, it's like finding the "slope" of a curve at a specific point. In the exercise, to solve for the derivative we are required to differentiate implicitly. This involves assuming that all terms involving the variable and its derivative exist independently within the equation, as in the equation given: \(xy^2 - 4x^{3/2} - y = 0\). Here we have two types of derivatives to compute. The straightforward derivatives like \( \frac{d}{dx}(-4x^{3/2}) \) which simplifies to \(-6x^{1/2}\), as well as implicit derivatives involving \( y \), where each term with \( y \) has its derivative considered with respect to \( x \). Knowing how to handle both types helps simplify and solve for the desired derivative function effectively.
Chain Rule
The chain rule is crucial when differentiating composite functions. It allows you to find the derivative of a function within a function, which occurs often in calculus. In implicit differentiation, the chain rule helps us when differentiating terms involving \( y \) because \( y \) is considered a function of \( x \). For example, in the term \( x y^2 \), you apply the chain rule to \( y^2 \), resulting in \( 2y \cdot \frac{dy}{dx} \). This will capture the rate of change of \( y \) with respect to \( x \). Remember that whenever you differentiate a term involving \( y \), you must multiply by \( \frac{dy}{dx} \) due to the chain rule.
Product Rule
The product rule is used when differentiating expressions where two functions are multiplied together. It states that the derivative of a product \( uv \) is \( u'v + uv' \). In this exercise, the term \( xy^2 \) requires the product rule because it is a product of \( x \) and \( y^2 \). Applying the rule:
  • Differentiate \( x \) to get \( 1 \), keeping \( y^2 \), resulting in \( y^2 \cdot 1 \).
  • Then differentiate \( y^2 \) to get \( 2y \cdot \frac{dy}{dx} \), and keep \( x \), resulting in \( 2xy \cdot \frac{dy}{dx} \).
Putting it together, we get \( y^2 + 2xy \cdot \frac{dy}{dx} \). Understanding when and how to apply the product rule allows for proper differentiation of complex terms as seen in this type of implicit differentiation problem.

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Most popular questions from this chapter

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1 .\) $$\begin{array}{|c|c|c|c|}\hline x & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 0 & {1} & {1} & {5} & {1 / 3} \\ \hline 1 & {3} & {-4} & {-1 / 3} & {-8 / 3} \\ \hline\end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x .\) $$\begin{array}{ll}{\text { a. } 5 f(x)-g(x),} & {x=1 \quad \text { b. } f(x) g^{3}(x), \quad x=0} \\ {\text { c. } \frac{f(x)}{g(x)+1}, \quad x=1} & {\text { d. } f(g(x)), \quad x=0}\end{array}$$ $$\begin{array}{l}{\text { e. } g(f(x)), \quad x=0 \quad \text { f. }\left(x^{11}+f(x)\right)^{-2}, \quad x=1} \\ {\text { g. } f(x+g(x)), \quad x=0}\end{array}$$

A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=x^{1 / 3}+x^{2 / 3}, \quad x_{0}=1$$

Tolerance The height and radius of a right circular cylinder are equal, so the cylinder's volume is \(V=\pi h^{3} .\) The volume is to be calculated with an error of no more than 1\(\%\) of the true Find approximately the greatest error that can be tolerated in the measurement of \(h\) , expressed as a percentage of \(h .\)

Moving along a parabola A particle moves along the parabola \(y=x^{2}\) in the first quadrant in such a way that its \(x\)-coordinate (measured in meters) increases at a steady 10 \(\mathrm{m} / \mathrm{sec} .\) How fast is the angle of inclination \(\theta\) of the line joining the particle to the origin changing when \(x=3 \mathrm{m} ?\)
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