Problem 22 - Calculus for Biology and Medicine

Question

Find the indicated partial derivatives. \(g(v, w)=\frac{w^{2}}{v+w} ; g_{v}(1,1)\)

Step-by-Step Solution

Answer

The partial derivative \(g_v(1,1) = -\frac{1}{4}\).

Step 1: Setting up the Partial Derivative

Identify which variable we need to differentiate with respect to. In this case, we are finding the partial derivative of the function \(g(v, w)\) with respect to \(v\), denoted as \(g_v(v, w)\).

Step 2: Applying the Quotient Rule

Notice that \(g(v, w) = \frac{w^2}{v+w}\) is a quotient. Use the Quotient Rule, which states that \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\), where \(u = w^2\) and \(v = v + w\).

Step 3: Differentiating the Numerator

Calculate the derivative of the numerator \(w^2\) with respect to \(v\). Since \(w^2\) is independent of \(v\), the derivative is 0: \(u' = 0\).

Step 4: Differentiating the Denominator

Calculate the derivative of the denominator \(v + w\) with respect to \(v\). The derivative is 1: \(v' = 1\).

Step 5: Evaluating the Quotient Rule

Substitute \(u'\), \(u\), \(v'\), and \(v\) into the Quotient Rule formula: \(g_v(v, w) = \frac{0(v + w) - w^2(1)}{(v + w)^2}\). This simplifies to \(g_v(v, w) = -\frac{w^2}{(v + w)^2}\).

Step 6: Substitute Values and Solve

Substitute \(v = 1\) and \(w = 1\) into the expression obtained in Step 5: \(g_v(1, 1) = -\frac{1^2}{(1 + 1)^2} = -\frac{1}{4}\).

Step 7: Conclusion

The partial derivative of \(g(v, w)\) with respect to \(v\), evaluated at \((1, 1)\), is \(-\frac{1}{4}\).

Key Concepts

Quotient Rule

The Quotient Rule is essential when dealing with problems involving division in differentiation. It's a formula that helps us find the derivative of a function that is a quotient of two differentiable functions. When you have a function in the form of \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of a variable (in this context, either \( v \) or \( w \)), the Quotient Rule says:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\]The intuition behind this rule is to carefully account for how both the numerator and the denominator change with respect to the variable considered.
It's crucial to memorize this formula and understand its application because it is a cornerstone in calculus when differentiating quotients.
Start by identifying each part correctly:

• \( u \) is the numerator expression
• \( v \) is the denominator expression

Once these are clear, calculate their derivatives. Substituting these into the Quotient Rule formula is the next step.
In our example, the function \( g(v, w) = \frac{w^2}{v+w} \) uses the Quotient Rule to find its partial derivative with respect to \( v \). This is a great demonstration of how the rule simplifies the calculations required.

Differentiation

Differentiation is a fundamental procedure in calculus. It involves computing the derivative of a function, which represents the rate of change. In partial derivatives, we take this concept and apply it to functions with multiple variables, focusing on one variable at a time.
By finding the partial derivative of a function like \( g(v, w) \), we observe how the function changes with respect to a single variable (either \( v \) or \( w \)), while treating the other variable as a constant.Let's break down the process:

• Identify the variable to differentiate with respect to. For \( g(v, w) \), we're interested in \( v \).
• Use differentiation rules like the Quotient Rule to proceed.
• Simplify the derivative to find the result.

In this case, because \( w^2 \) is independent of \( v \), its derivative is \( 0 \), which greatly simplifies the problem.
Understanding differentiation principles is critical because it forms the backbone of calculus, enabling us to model and solve real-world problems.

Numerator and Denominator Derivatives

In problems involving the Quotient Rule, correctly finding the derivatives of the numerator and denominator functions is crucial. These derivatives are pivotal, as they directly influence the result of the quotient's differentiation.

Numerator Derivative

The numerator in our function, \( u = w^2 \), is independent of \( v \). Therefore, when calculating the partial derivative with respect to \( v \), \( u' \) is \( 0 \). This happens because \( w^2 \) doesn't change as \( v \) changes.
It's important to identify such cases because they simplify computations significantly.

Denominator Derivative

For the denominator, \( v = v + w \), the derivative with respect to \( v \) is \( 1 \). Here, \( w \) is treated as a constant, and the derivative of \( v \) is straightforward.
These derivatives feed into the Quotient Rule formula, providing the components needed for the final calculation.By understanding and efficiently calculating these derivatives, you simplify the process of finding the partial derivative using the Quotient Rule. Correct application results in more accurate and effective problem solving.