Problem 17 - Calculus an Applied Approach

Question

Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the given point. Equation \(\quad\) Point \(x^{3}-x y+y^{2}=4 \quad(0,-2)\)

Step-by-Step Solution

Answer

The derivative of the function at the point (0,-2) is 1.

Step 1: Apply the implicit differentiation

Applying the derivative to both sides of the equation \(x^{3}-x y+y^{2}-4=0\), on LHS, the derivative of \(x^3\) with respect to x is \(3x^2\), the derivative of \(-xy\) with respect to x is \(-y - x \frac{dy}{dx}\) using the product rule, the derivative of \(y^2\) with respect to x is \(2y \frac{dy}{dx}\) using the chain rule, and the derivative of -4 is 0. So, d/dx[LHS] results in \(3x^{2}-y-x \frac{dy}{dx}+2y \frac{dy}{dx}=0\). Keep this equation for later use.

Step 2: Simplify the derivative obtained

Now we simplify the equation we got in Step 1, \(3x^{2}-y-x \frac{dy}{dx}+2y \frac{dy}{dx}=0\), by moving non-derivative terms to the right and collecting all the derivative terms. You obtain: \(x \frac{dy}{dx}-2y \frac{dy}{dx}=y-3x^{2}\), then factor out \(\frac{dy}{dx}\) to isolate it on one side: \(\frac{dy}{dx}(x-2y)=y-3x^{2}\). Finally you solve for \(\frac{dy}{dx}\) to get the answer: \(\frac{dy}{dx}=\frac{y-3x^{2}}{x-2y}\).

Step 3: Evaluate the derivative at the given point

Next, you substitute the values from the point (0,-2) into the derived expression for \(\frac{dy}{dx}\). Hence, \(\frac{dy}{dx}=\frac{-2-3(0^{2})}{0-2(-2)}\). A simplification of this will give \(\frac{dy}{dx}=1\).

Key Concepts

Product Rule

The Product Rule is a fundamental technique in calculus used to differentiate expressions where two functions are multiplied together. It's essential for solving implicit differentiation problems, where each term might involve multiple variables. The Product Rule states:

• If you have two functions, say, \( u(x) \) and \( v(x) \), the derivative of their product \( u(x) \, \cdot \, v(x) \) is given by: \( u'v + uv' \). Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

In the context of implicit differentiation, this rule helps tackle terms like \(-xy\) in our exercise. We treat \(x\) and \(y\) as separate functions of \(x\). Thus, applying the Product Rule to \(-xy\), you get: \(-y - x\frac{dy}{dx}\). The \(-y\) comes from differentiating \(x\) (due to \(y\) being constant momentarily), and \(-x\frac{dy}{dx}\) comes from differentiating \(y\) as a function of \(x\).
This way, the Product Rule facilitates expressing complex terms in dynamic equations.

Chain Rule

The Chain Rule is another pivotal calculus tool that allows us to differentiate compositions of functions. It's especially useful in implicit differentiation when functions are intertwined, and variables depend on each other implicitly. The rule can be stated as:

• For any function \( g(f(x)) \), its derivative is \( g'(f(x)) \cdot f'(x) \).

In our exercise, this rule is crucial when differentiating terms involving squared variables like \( y^2 \). Here, \( y \) is assumed to be implicitly dependent on \( x \). If \( f(y) = y^2 \), using the Chain Rule gives the derivative as: \( 2y \cdot \frac{dy}{dx} \).
This highlights how the Chain Rule helps manage complex dependencies between variables. By providing a way to elegantly handle these internal relationships, it becomes indispensable when tackling implicit differentiation problems.

Calculus Problem Solving

Solving calculus problems often involves breaking down complex tasks into more manageable steps using various rules and techniques. Implicit differentiation itself is centered around this process, providing a method to find derivatives of functions defined implicitly rather than explicitly.Such problems typically include:

• Identifying the Equation: Recognizing the relationship between variables.
• Applying the Right Rules: Using Product or Chain Rules where relevant to deal with function compositions and products.

For the given exercise, solving involves:

• Careful Differentiation: Implicitly differentiate every term in the equation, often requiring a keen eye to correctly apply differentiation rules.
• Equation Simplification: Rearranging and simplifying the derived equation to solve for \( \frac{dy}{dx} \).

Each problem-solving process in calculus bolsters the understanding and application of mathematical theory, turning abstract concepts into concrete solutions.

Evaluate Derivative at a Point

After deriving a general formula for \( \frac{dy}{dx} \), evaluating it at a specific point gives the slope of the tangent line at that point. It's like zooming in on one part of the curve, offering insights into its characteristics such as steepness or direction.The process involves:

• Substitution: Plugging the coordinates of the given point into the derivative expression.

For our exercise, once \( \frac{dy}{dx} \) was calculated as \( \frac{y-3x^2}{x-2y} \), the point \( (0, -2) \) was inserted into this formula. Therefore, the derivative simplified to \( 1 \) after substitution.
This tells us that at the point \( (0, -2) \), the curve is rising with a slope of \(1\), showing the power of derivatives in providing numerical insights into graphs and real-world modeling.