Problem 24 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ x y^{2}-4 x^{3 / 2}-y=0 $$

Step-by-Step Solution

Answer

\( dy = \frac{6x^{1/2} - y^{2}}{2xy - 1} \cdot dx \)

Step 1: Differentiate with respect to x

To find \(dy\), we need to differentiate the given equation implicitly with respect to \(x\). The equation is \(x y^{2} - 4x^{3/2} - y = 0\). Differentiating each term with respect to \(x\): 1. \(\frac{d}{dx}(x y^{2}) = y^{2} + 2xy \frac{dy}{dx}\) (using the product rule) 2. \(\frac{d}{dx}(-4x^{3/2}) = -6x^{1/2}\) (using power rule) 3. \(\frac{d}{dx}(-y) = -\frac{dy}{dx}\) (with respect to \(y\))Combine these results to get the differentiated equation:\[ y^{2} + 2xy \frac{dy}{dx} - 6x^{1/2} - \frac{dy}{dx} = 0 \]

Step 2: Collect terms involving dy/dx

Rearrange the terms to isolate the \(\frac{dy}{dx}\) on one side of the equation. Combine like terms:\[ 2xy \frac{dy}{dx} - \frac{dy}{dx} = 6x^{1/2} - y^{2} \]Simplify the left side by factoring out \(\frac{dy}{dx}\): \[ (2xy - 1) \frac{dy}{dx} = 6x^{1/2} - y^{2} \]

Step 3: Solve for dy/dx

To find \(\frac{dy}{dx}\), divide both sides of the equation by the factor \((2xy - 1)\): \[ \frac{dy}{dx} = \frac{6x^{1/2} - y^{2}}{2xy - 1} \]

Step 4: Express dy as a differential

Finally, express \(dy\) as a differential in terms of \(dx\): \[ dy = \frac{6x^{1/2} - y^{2}}{2xy - 1} \cdot dx \]

Key Concepts

Product Rule

In differential calculus, the product rule is essential for differentiating the product of two functions. When you have a function, say, \( u(x)v(x) \), the rule states that:

• \( \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) \).

This rule is used in the given exercise to differentiate the term \( xy^2 \).
To apply the product rule:

• Let \( u = x \) and \( v = y^2 \).
• The derivative \( u'(x) = 1 \).
• The derivative \( v'(x) = 2y \frac{dy}{dx} \), using the chain rule here.
• Thus, the differentiated term is \( y^2 + 2xy \frac{dy}{dx} \).

This result helps in managing complex equations where both functions and their variables are intertwined.

Power Rule

The power rule is a foundational concept used to differentiate terms involving powers of \( x \). It states:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)

For any function of \( x \), where \( n \) is a real number, the power rule simplifies finding derivatives.
In our exercise, it's applied to \(-4x^{3/2}\):

• Using the power rule, differentiate \( -4x^{3/2} \):
• \( -4 \cdot \frac{3}{2} \cdot x^{1/2} = -6x^{1/2} \).

This concept is crucial as it allows us to differentiate polynomial terms efficiently. By applying the power rule, calculations become straightforward and help in solving differential equations.

Differential Calculus

Differential calculus focuses on the concept of a derivative, which represents the rate of change of a function. In essence, it's about how a function changes when its input changes.
Key aspects of differential calculus include:

• Understanding derivatives as slopes of tangents to a curve.
• Using rules like product and power rules for differentiating functions.
• Applying these rules to find the derivative of implicit functions, which don’t isolate \( y \) explicitly.

In our problem, differential calculus helps us determine \( \frac{dy}{dx} \) by implicitly differentiating the given equation. This transforms a multifaceted equation into something more manageable, aiding in understanding how variables interact in multi-variable functions.

Implicit Function Theorem

The implicit function theorem is a significant tool in calculus used for functions defined implicitly rather than explicitly. Here’s what you need to know:

• It allows differentiation of functions even when they are not solved for one variable.
• This theorem is vital when equations have no simple algebraic rearrangement.
• It helps us derive \( \frac{dy}{dx} \) directly, as seen in our exercise.

In implicit differentiation, you treat \( y \) as a function of \( x \) even if \( y \) is not given explicitly as \( y = f(x) \).
For our example, the equation \( x y^2 - 4 x^{3/2} - y = 0 \) doesn't solve easily for \( y \). By differentiating each term w.r.t \( x \) and using related calculus rules, we derive relations between derivatives. This approach demonstrates the power of implicit differentiation in solving equations involving multiple variables and functions.