Problem 63 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative of the function is \(f'(x) = 3x^2 - \frac{3}{x^4}\).

Step 1: Apply the power rule to the first term

The power rule for differentiation states that if you have a term of the form \(x^n\), the derivative is \(nx^{n-1}\). Apply this to the term \(x^3\). The derivative of \(x^3\) is \(3x^{3-1} = 3x^2\).

Step 2: Convert the second term to a power form

Before differentiating the second term, convert \(\frac{1}{x^3}\) into a form that uses exponents. This can be rewritten as \(x^{-3}\).

Step 3: Apply the power rule to the converted second term

Now apply the power rule to the term \(x^{-3}\). The derivative is \(-3x^{-3-1} = -3x^{-4}\).

Step 4: Combine the derivatives

Combine the results from Steps 1 and 3 to get the complete derivative. Thus, \(f'(x) = 3x^2 - 3x^{-4}\).

Step 5: Simplify the result

Express \(-3x^{-4}\) as \(-\frac{3}{x^4}\). Combine everything to get a clean expression for the derivative: \(f'(x) = 3x^2 - \frac{3}{x^4}\).

Key Concepts

Power Rule of Differentiation

The power rule is one of the most fundamental tools for finding derivatives, specifically for functions involving powers of a variable. It guides us to efficiently find the derivative by manipulating the exponent of each term in the function. When you see a term in the form of \(x^n\), its derivative can be easily calculated using the power rule formula: \[ \frac{d}{dx}x^n = nx^{n-1} \] This means you take the exponent \(n\), multiply it by the coefficient of \(x\), and then reduce the exponent by one. This nifty little rule helps to quickly tackle polynomial functions. For example, if you have \(x^3\), applying the power rule gives you \(3x^2\), making the process straightforward and manageable. Uses of the power rule include: - Simplifying the differentiation of polynomial functions - Serving as the base for understanding more complex calculus concepts Remember, the power rule is applicable only to terms where the variable \(x\) is raised to a power, including integer and fractional exponents.

Understanding Derivatives

Derivatives are a key concept in calculus, providing insights into the rate of change of a function. When you differentiate a function, you essentially find how the function's output changes as its input changes. This concept is critical in fields such as physics, engineering, and economics, where change and variation are constant factors. Key points about derivatives: - Derivatives signify instantaneous rate of change at any point of a function - They are symbolically represented as \(f'(x)\) or \(\frac{df}{dx}\) for function \(f\) with respect to variable \(x\) Consider function \(f(x) = x^3 + \frac{1}{x^3}\). By differentiating it, using rules like the power rule, you learn about the slopes and behavior of this polynomial function at any given point. Derivative applications are vast, allowing you to find: - Slopes of tangent lines - Velocities and accelerations in physics - Optimal solutions in economic models

Exponent Rules

Understanding exponent rules is crucial when dealing with differentiation, especially when terms initially appear in forms like fractions or require exponent manipulation. Exponent rules allow you to rewrite terms in a form that makes differentiation straightforward and simple.Key exponent rules to remember: - \(x^{-n} = \frac{1}{x^n}\), which transforms reciprocal functions into exponential form for easier differentiation - \(x^0 = 1\), maintaining consistency in calculations For example, in the function \(f(x) = \frac{1}{x^3}\), applying the rule \(x^{-n} = \frac{1}{x^n}\) helps you express it as \(x^{-3}\). This conversion is crucial because it allows you to directly apply the power rule to find the derivative. Without a strong grasp of exponent rules, differentiating functions composed of fractional and negative exponents becomes tedious. Mastering these rules enhances your ability to handle various calculus problems effortlessly, leading to accurate and efficient solutions.