Problem 22 - Calculus for Biology and Medicine

Question

In Problems 17-24, 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 \( g_v(1, 1) \) is \(-\frac{1}{4}\).

Step 1: Understand the Function

We are given a function \( g(v, w) = \frac{w^2}{v+w} \) and we need to find the partial derivative of \( g \) with respect to \( v \) at the point \((v, w) = (1, 1)\).

Step 2: Apply the Quotient Rule

The function is of the form \( \frac{u}{v} \) where \( u = w^2 \) and \( v = v + w \). The quotient rule states that the derivative of \( \frac{u}{v} \) is \( \frac{v \cdot u' - u \cdot v'}{v^2} \).

Step 3: Find Derivatives of Numerator and Denominator

First, find the partial derivative of the numerator \( u = w^2 \) with respect to \( v \), which is 0 because \( u \) does not depend on \( v \). Next, find the partial derivative of the denominator \( v + w \) with respect to \( v \), which is 1.

Step 4: Substitute into the Quotient Rule

Substitute these derivatives into the quotient rule: \[ g_v(v, w) = \frac{(v+w) \cdot 0 - w^2 \cdot 1}{(v+w)^2} = \frac{-w^2}{(v+w)^2} \]

Step 5: Evaluate at the Given Point

Now, we evaluate \( g_v(v, w) \) at \( v = 1 \) and \( w = 1 \): \[ g_v(1, 1) = \frac{-(1^2)}{(1+1)^2} = \frac{-1}{4} \]

Key Concepts

Quotient Rule

When working with functions that are ratios of two expressions, the quotient rule is essential for differentiating such functions. If you have a function denoted by \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of the variables involved, the quotient rule provides a structured way to find the derivative of this entire function.

The quotient rule formula is:

• \( \frac{v \cdot u' - u \cdot v'}{v^2} \)

Here's what each component represents:

• \( u' \): the derivative of the numerator \( u \)
• \( v' \): the derivative of the denominator \( v \)
• \( v^2 \): the square of the original denominator \( v \)

By subtracting the product of the numerator's derivative and denominator's derivative scaled by the inverse square of the denominator, we account for the way changes in both \( u \) and \( v \) affect the quotient as a whole. This tool becomes particularly powerful when dealing with partial derivatives, like in our given problem.

Numerator and Denominator Derivatives

To effectively employ the quotient rule, we need to carefully determine the derivatives of both the numerator and the denominator in the function. In the function given in the original problem, \( g(v, w) = \frac{w^2}{v+w} \), the numerator is \( u = w^2 \), and the denominator is \( v + w \).

For partial derivatives, examine how each component changes with respect to a single variable, in this case, \( v \):

• **Numerator Derivative**: Determine the partial derivative of \( u = w^2 \) with respect to \( v \). Since \( u \) is solely a function of \( w \), its partial derivative with respect to \( v \) is 0.
• **Denominator Derivative**: Determine the partial derivative of \( v + w \) with respect to \( v \). Here, only \( v \) is differentiable with respect to itself, resulting in a derivative of 1.

A clear understanding of these derivatives supports correct application of the quotient rule. Thus ensuring smooth calculation of the partial derivative, leading to the final result.

Evaluating Derivatives at a Point

Once the partial derivative expression is calculated using the quotient rule and the component derivatives, the next step is evaluating it at a specific point. This refines a broad formula into the precise value needed for analysis.

Given our example \( g_v(v, w) = \frac{-w^2}{(v+w)^2} \), it's crucial to substitute \( v = 1 \) and \( w = 1 \) to find the exact partial derivative at this point. Upon making these substitutions:

• The numerator transforms into \( -1 \) given \( w = 1 \).
• The denominator simplifies to \( 4 \) from \( (1+1)^2 \).

This results in:

• \( g_v(1, 1) = \frac{-1}{4} \)

Thus, determining the value of the partial derivative at a specific point involves precise substitution, leading to a succinct answer that reflects changes in \( g \) with respect to \( v \) at that particular coordinate.