Problem 24 - University Calculus: Early Transcendentals

Question

Find \(d y\). $$x y^{2}-4 x^{3 / 2}-y=0$$

Step-by-Step Solution

Answer

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

Step 1: Differentiate both sides with respect to x

We start by differentiating the equation \( xy^2 - 4x^{3/2} - y = 0 \) with respect to \( x \). This requires using the product rule for \( xy^2 \) and the chain rule for the other terms.- Differentiate \( xy^2 \): \( \frac{d}{dx}(xy^2) = x \frac{d}{dx}(y^2) + y^2 \frac{d}{dx}(x) = 2xy \frac{dy}{dx} + y^2 \).- Differentiate \( 4x^{3/2} \): \( \frac{d}{dx}(4x^{3/2}) = 6x^{1/2} \).- Differentiate \( y \): \( \frac{d}{dx}(y) = \frac{dy}{dx} \).

Step 2: Set up the differentiated equation

Substitute the derived derivatives from Step 1 into the original equation:\[2xy \frac{dy}{dx} + y^2 - 6x^{1/2} - \frac{dy}{dx} = 0\]This is now an equation involving \( \frac{dy}{dx} \) which can be rearranged to solve for \( \frac{dy}{dx} \).

Step 3: Rearrange the equation to isolate dy/dx

Group the terms involving \( \frac{dy}{dx} \) together:\[(2xy - 1) \frac{dy}{dx} = 6x^{1/2} - y^2\]

Step 4: Solve for dy/dx

Solve the equation from Step 3 for \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{6x^{1/2} - y^2}{2xy - 1}\]

Step 5: Verify the solution

Check that substituting back into the differentiated equation satisfies the conditions for differentiation of the original equation. If correct, the solution for \( \frac{dy}{dx} \), which represents \( dy \), is complete.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It's the process of finding the derivative of a function, which represents the rate at which a function is changing at any given point. To differentiate a function means to compute its derivative, denoted as \( \frac{dy}{dx} \) for a function \( y \) with respect to \( x \). The derivative tells us how the function behaves and helps in understanding the slopes and rates of change.

• For example, if \( y \) is a function of \( x \), the derivative \( \frac{dy}{dx} \) gives the rate of change of \( y \) with respect to \( x \).
• A derivative can show whether a function is increasing or decreasing, and it tells us the steepness of the curve at any point.

Differentiation requires familiarity with rules for derivatives for common functions, such as polynomials, trigonometric, exponential, and logarithmic functions. Understanding these rules is pivotal for solving calculus problems effectively.

Product Rule

The product rule is a calculus principle used when differentiating products of two or more functions. If you have two functions, \( u(x) \) and \( v(x) \), and you wish to differentiate their product \( u(x)v(x) \), the product rule states:
\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] This means you take the derivative of the first function and multiply it by the second function, then take the derivative of the second function and multiply it by the first. Add these two results together.

• In the context of the exercise, the term \( xy^2 \) requires the use of the product rule: Differentiating \( x \) and \( y^2 \) individually and combining them results in \( 2xy \frac{dy}{dx} + y^2 \).
• It simplifies the differentiation of complex functions and shows the interplay between two multiplied functions.

The product rule is an invaluable tool in calculus for dealing with functions that are products of one another, making it easier to tackle challenging differentiation problems.

Chain Rule

The chain rule is another essential tool in calculus. It's used when differentiating composite functions, ensuring that each layer of the function is properly accounted for. If you have a composite function \( f(g(x)) \), the chain rule allows us to find its derivative by following these steps:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \, g'(x) \] This involves differentiating the outer function \( f \) and multiplying by the derivative of the inner function \( g \).

• For instance, in differentiating \( y^2 \), the chain rule helps in recognizing that \( y \) is a function of \( x \), so we take the derivative with respect to \( y \) and then multiply it by \( \frac{dy}{dx} \).
• This rule simplifies the process of finding derivatives of complex expressions, which might otherwise be difficult to manage.

The chain rule is particularly important when functions are nested or involve other variable expressions, enhancing problem-solving in advanced calculus.

Implicit Differentiation

Implicit differentiation is a powerful technique used when functions are not given in a conventional \( y = f(x) \) form. It is often applied to relations where \( y \) and \( x \) are interdependent and intertwined, making it impossible to easily solve for \( y \) in terms of \( x \) before differentiating.
Here's how implicit differentiation works:

• Differentiate every term of the equation with respect to \( x \), recognizing that \( y \) is a function of \( x \).
• Whenever a \( y \) term is differentiated, include \( \frac{dy}{dx} \) to capture the change in \( y \) with respect to \( x \).

In the original exercise, implicit differentiation lets us tackle the equation \( xy^2 - 4x^{3/2} - y = 0 \), by treating \( y \) as a function of \( x \) throughout.
This method unlocks solutions for complex equations and is crucial when direct differentiation isn't feasible. It's a core skill in calculus, helping to find derivatives when dealing with linked variables.