Problem 30 - Calculus for Biology and Medicine

Question

Differentiate $$ f(x)=\frac{r+x}{r s^{2}}-r s x+(r+s) x-r s $$

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{rs^2} - rs + r + s\).

Step 1: Rewrite the Function

First, we start by rewriting the function for clarity: \[ f(x) = \frac{r + x}{r s^{2}} - r s x + (r + s) x - r s \] This function consists of four terms. We'll differentiate each term separately applying the rules of differentiation.

Step 2: Differentiate the First Term

The first term is \((r+x) / (rs^2)\). This can be rewritten as \(\frac{1}{rs^2}(r + x)\). The derivative is found by applying the constant and linear differentiation rules:\[\frac{d}{dx}\left(\frac{r+x}{rs^2}\right) = \frac{1}{rs^2} \cdot \frac{d}{dx}(r + x) = \frac{1}{rs^2} \cdot (0 + 1) = \frac{1}{rs^2}.\]

Step 3: Differentiate the Second Term

The second term is \(-rsx\). Applying the power rule gives:\[\frac{d}{dx}(-rsx) = -rs \cdot 1 = -rs.\]

Step 4: Differentiate the Third Term

The third term is \((r + s)x\). Applying the power rule results in:\[\frac{d}{dx}((r+s)x) = (r+s) \cdot 1 = r+s.\]

Step 5: Differentiate the Fourth Term

The fourth term, \(-rs\), is a constant. The derivative of a constant is 0:\[\frac{d}{dx}(-rs) = 0.\]

Step 6: Combine the Derivatives

Add together the derivatives of each term to get the derivative of the entire function:\[\frac{d}{dx}f(x) = \frac{1}{rs^2} - rs + (r+s) + 0 = \frac{1}{rs^2} - rs + r + s.\]

Key Concepts

Constant Differentiation

When we differentiate a term that includes both variables and constants, it is important to understand how each component affects the outcome. In mathematics, a constant is a fixed value that does not change. Constants don't have variables attached to them, like \( r \) and \( s \) in our given function.

In differentiation, whenever we apply the derivative to a constant by itself, its derivative is zero. This is because a constant does not change regardless of the value of the variable(s) in the function. No slope can be assigned because there's no change.

• The derivative of \(-rs\) in our example yields \(0\) because it is simply a constant.

By differentiating constants correctly, we maintain the accuracy of our results.

Linear Differentiation

Linear functions take on the form \(y = mx + c\). They are straightforward to differentiate. The term involving linear differentiation in this problem is \((r + x)/(rs^2)\). When we differentiate this expression, we see that:

• The \(x\) component results in a value, as \( \frac{d}{dx}(x) = 1 \).
• Terms like \( r \), without an \(x\) following them, are constants, thus their contribution upon differentiation is zero.

For linear functions, the slope (or derivative) is constant. We simply apply the linear differentiation rule to get the required result.

Power Rule

Among the most common rules in differentiation, the power rule allows for quick calculations. The rule states: if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). It applies to powers of \(x\), indicating how these terms evolve when differentiated.

In our function, terms like \(-rsx\) utilize the power rule. With \(n=1\), differentiating results in \(-rs \cdot 1 = -rs\), indicating a slope guided by constant multiplication.

• The same approach goes for \((r+s)x\), where differentiation gives \(r+s\), due to \(x\) being raised to the first power.

Mastering the power rule aids in maneuvering through more complex terms with ease.

Derivative of a Constant

Understanding that the derivative of a constant is zero is fundamental. A constant's derivative reflects the lack of dependency on a variable.

In our problem, \(-rs\) is a term solely made up of constants, and so differentiating with respect to \(x\) yields zero:

• This principle applies universally, regardless of specific values.
• This influences our function by removing unchanging components.

Grasping this concept enables quick assessments and efficient differentiation of mathematical functions that integrate constants.