Problem 23 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ y=\left(\frac{1}{w^{3}-1}\right)^{4} $$

Step-by-Step Solution

Answer

The derivative of the function is \(-\frac{12w^2}{(w^3-1)^5}\).

Step 1: Identify the Outer and Inner Functions

The function given is a composition of two functions: \(y = (u)^4\) where \(u = \frac{1}{w^3 - 1}\). Consider \(f(u)=u^4\) as the outer function and \(g(w)=\frac{1}{w^3 - 1}\) as the inner function.

Step 2: Differentiate the Outer Function

Use the power rule to differentiate the outer function. If \(u^n\), its derivative is \(n\cdot u^{n-1}\). Thus, the derivative of \(f(u)=u^4\) is \(f'(u)=4u^3\). String this together with the inner derivative using the chain rule.

Step 3: Differentiate the Inner Function

Differentiate the inner function \(g(w) = \frac{1}{w^3 - 1}\). Apply the quotient rule: \(h'(x) = \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). Here, \(u = 1\) and \(v = w^3 - 1\). So, \(u' = 0\) and \(v' = 3w^2\). The derivative \(g'(w)= -\frac{3w^2}{(w^3 - 1)^2}\).

Step 4: Apply the Chain Rule

Now combine the derivatives using the chain rule: \(\frac{dy}{dw}=f'(g(w))\cdot g'(w)\). Substitute to get \(\frac{dy}{dw} = 4\left(\frac{1}{w^3 - 1}\right)^3 \cdot \left(-\frac{3w^2}{(w^3 - 1)^2}\right)\).

Step 5: Simplify the Expression

Simplify the expression: \[\frac{dy}{dw} = -\frac{12w^2}{(w^3 - 1)^5}.\] This is your final answer for the derivative of the given function.

Key Concepts

Chain Rule

When dealing with composite functions, such as one inside another, the chain rule is a powerful tool for finding derivatives. This rule essentially lets you "chain" the derivatives of these nested functions together.
To apply the chain rule, follow these steps:

• Identify the outer and inner functions. In this case, our outer function is \(f(u) = u^4\) and the inner function is \(g(w) = \frac{1}{w^3 - 1}\).
• Find the derivative of the outer function using the outer function value as the variable. Here, we'd differentiate \(f(u) = u^4\) to get \(f'(u) = 4u^3\).
• Next, take the derivative of the inner function with respect to its variable. Thus, differentiating \(g(w)\) gives us \(g'(w) = -\frac{3w^2}{(w^3 - 1)^2}\).
• Finally, use the chain rule to combine these results: \(\frac{dy}{dw} = f'(g(w)) \cdot g'(w)\).

By applying this method, you can easily differentiate composite functions, even if they look complicated at first glance.

Power Rule

The power rule is one of the first derivative rules learned in calculus, and it plays a critical role in differentiating polynomials. The rule states that for any function of the form \(u^n\), its derivative is \(n \cdot u^{n-1}\).
This rule simplifies the process of taking derivatives of powers of a variable. For example, when attempting to find the derivative of \(f(u) = u^4\), the power rule tells us that the derivative is \(f'(u) = 4u^3\). We simply multiply the coefficient (4) by the variable raised to the power one less than the original exponent (3).
Hence, the power rule frees you from complex differentiation processes for functions that fit its pattern, saving time and reducing potential mistakes in manual calculation.

Quotient Rule

The quotient rule allows us to differentiate functions that are divided by another. For a function \(g(w) = \frac{u}{v}\), its derivative is given by the formula:

• \(h'(w) = \frac{u'v - uv'}{v^2}\)

Here's how to apply it:

• Identify the numerator (\(u\)) and the denominator (\(v\)) functions. In this instance, \(u = 1\) and \(v = w^3 - 1\).
• Calculate \(u'\) and \(v'\). Since \(u\) is a constant, \(u' = 0\). Meanwhile, \(v'\) is derived from \(v\), yielding \(3w^2\).
• Substitute these into the quotient rule formula, resulting in \(g'(w) = -\frac{3w^2}{(w^3 - 1)^2}\).

Understanding the quotient rule is crucial when working with ratios, enabling us to work with more complex fractions and their derivatives.