Problem 66 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. \(f(x)=-x^{3}+\frac{2 x^{2}+3}{x^{4}+1}\)

Step-by-Step Solution

Answer

The derivative of \( f(x) \) is \( f'(x) = -3x^2 + \frac{-4x^5 - 12x^3 + 4x}{(x^4 + 1)^2} \).

Step 1: Differentiate the first term

The function is given as \( f(x) = -x^3 + \frac{2x^2 + 3}{x^4 + 1} \). First, we'll differentiate \(-x^3\) with respect to \(x\). The derivative of \(-x^3\) is given by using the power rule, \( \frac{d}{dx}(-x^3) = -3x^2 \).

Step 2: Apply the quotient rule

For the term \( \frac{2x^2 + 3}{x^4 + 1} \), we use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \). Let \( u = 2x^2 + 3 \) and \( v = x^4 + 1 \).

Step 3: Differentiate the numerator

Differentiate \( u = 2x^2 + 3 \) with respect to \(x\). The derivative is \( \frac{du}{dx} = 4x \).

Step 4: Differentiate the denominator

Differentiate \( v = x^4 + 1 \) with respect to \(x\). The derivative is \( \frac{dv}{dx} = 4x^3 \).

Step 5: Substitute into the quotient rule formula

Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula: \( \frac{d}{dx} \left( \frac{2x^2 + 3}{x^4 + 1} \right) = \frac{(x^4 + 1)(4x) - (2x^2 + 3)(4x^3)}{(x^4 + 1)^2} \). Simplify the expression.

Step 6: Simplify the expression

Expanding: \( (x^4 + 1)(4x) = 4x^5 + 4x \) and \( (2x^2 + 3)(4x^3) = 8x^5 + 12x^3 \). Thus, the numerator becomes \( 4x^5 + 4x - 8x^5 - 12x^3 = -4x^5 - 12x^3 + 4x \).

Step 7: Combine results

Now, combine the results from Steps 1 and 6: \( f'(x) = -3x^2 + \frac{-4x^5 - 12x^3 + 4x}{(x^4 + 1)^2} \).

Key Concepts

Power Rule

The Power Rule is an essential tool in calculus, especially when dealing with polynomials. It states that if you have a term like \( x^n \), where \( n \) is a constant, the derivative is \( nx^{n-1} \). Simply put, you multiply the power by the coefficient, then decrease the power by one. For example, in the term \(-x^3\), using the Power Rule:

• First, multiply the power (3) by the coefficient (-1) to get \(-3\).
• Then, reduce the power by one, resulting in \(-3x^2\).

The application of this rule makes it simpler to differentiate each term in a polynomial expression efficiently.

Quotient Rule

The Quotient Rule is used in differentiation when you have a function that is the division of two other functions, say \( f(x) = \frac{u}{v} \). To find the derivative of such a function, the Quotient Rule formula comes into play: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \] Breaking it down:

• First, differentiate the numerator \( u \) and the denominator \( v \), giving you \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).
• Substitute these derivatives back into the quotient formula, which calculates the overall change of the ratio.
• Finally, this needs to be simplified, which can sometimes involve expanding and combining like terms.

The Quotient Rule is particularly useful in more complex expressions where division is involved, enabling precise derivation.

Calculus in Biology

Calculus plays an invaluable role in biology by helping in the analysis of dynamic systems. For instance, it models how populations change over time given birth and death rates. Derivatives are used to find the rate of change of population size or other biological processes. Examples include:

• Tracking the growth rate of bacteria in a culture, where calculus provides the relationship between time and population size.
• Determining the rate at which a drug is metabolized in the body, helping in pharmacokinetics.
• Understanding enzyme activity rates, which are essential in biochemistry and medicine.

By applying calculus and derivatives, biologists can make informed predictions about natural processes, aiding research and application in various biological fields.

Mathematical Derivatives

Mathematical derivatives are critical in understanding how functions change. A derivative gives you the rate of change of a function's value relative to changes in its input. In simple terms, it measures how a function behaves as its variables change. Derivatives have multiple applications:

• Used in physics to determine velocity and acceleration of objects, where calculus is needed to describe motion.
• In economics, derivatives help to assess cost, revenue, or profit functions based on varying production levels.
• In engineering, they assist in designing systems and understanding physical phenomena.

Understanding derivatives is fundamental for any calculus student, as they form the basis for more advanced mathematical concepts and applied mathematics scenarios.