Problem 9 - Calculus for Biology and Medicine

Question

In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=\frac{1}{\left(x^{3}-1\right)^{4}} $$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Form

The given function is\[f(x) = \frac{1}{(x^3 - 1)^4}.\]This is essentially a power function written as a fraction which can be transformed using the power rule and the chain rule.

Step 2: Rewrite the Function

Rewrite the function in a form that's easier to differentiate:\[f(x) = (x^3 - 1)^{-4}.\]

Step 3: Differentiate Using the Chain Rule

Apply the chain rule to differentiate. The chain rule states\[\frac{d}{dx}[u^n] = nu^{n-1}\cdot\frac{du}{dx},\]where \(u(x) = x^3 - 1\) and \(n = -4\).

Step 4: Find the Derivative of the Inner Function

Differentiate the inner function \(u(x) = x^3 - 1\):\[u'(x) = 3x^2.\]

Step 5: Apply the Chain Rule

Now apply the chain rule to differentiate the outer function:\[f'(x) = -4(x^3 - 1)^{-5} \cdot 3x^2.\]

Step 6: Simplify the Derivative Expression

Simplify the expression:\[f'(x) = -12x^2(x^3 - 1)^{-5}.\]

Key Concepts

Understanding the Chain Rule

The chain rule is a powerful technique in differentiation, particularly useful when dealing with composite functions. A composite function is a function that is applied to another function. When you have a function nested within another function, the chain rule allows you to find the derivative effectively.

The chain rule states that if you have a function of the form \(f(g(x))\), where \(f\) is the outer function and \(g\) is the inner function, then the derivative \(f'(x)\) is found by multiplying the derivative of the outer function by the derivative of the inner function.

In mathematical terms:

• Let \(y = f(u)\) where \(u = g(x)\),
• Then \(\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx}\).

In the example problem, \(u = x^3 - 1\) is the inner function, and the outer function is \(u^{-4}\). First, differentiate the inner function, then apply the derivative to the outer function, and multiply them together.

This method is efficient and simplifies complex differentiation tasks.

Exploring the Power Rule

The power rule is a fundamental tool in calculus used to differentiate functions of the form \(x^n\), where \(n\) is any real number. This rule makes it easy to manage polynomials and expressions involving powers.

According to the power rule, the derivative of \(x^n\) is given by:

\[\frac{d}{dx}[x^n] = nx^{n-1}\] The power rule simplifies the process by allowing you to multiply the exponent by the coefficient and reduce the power by one.

In the problem provided, the function was rewritten as \((x^3 - 1)^{-4}\). The power rule is applied directly to handle the \(-4\) exponent, allowing us to multiply this by the derivative of the inner function (found using the chain rule).

This method helps break down complex differentiations into manageable steps, especially when combined with the chain rule.

Mastering Calculus Problem-Solving

Solving calculus problems effectively involves understanding key strategies like identifying function types and applying the correct differentiation rules. In our example, the problem starts with identifying that the original form \(\frac{1}{(x^3 - 1)^4}\) is a function suitable for the power and chain rules.

Here are some steps to follow in calculus problem-solving:

• Recognize the form: Understand the type of function you're working with. Is it a polynomial, a composite function, a product, or a quotient?
• Rewrite the function: Simplify the expression to make differentiation easier. Functions can often be rewritten in a more manageable form.
• Apply appropriate rules: Use rules like the chain rule and power rule to find the derivative. Often, more than one rule is necessary.
• Verify and simplify: After differentiating, simplify the result and check your work for accuracy.

By following these strategies, students can achieve a more structured approach to solving calculus problems, leading to a deeper understanding and mastery of calculus concepts.