Problem 13 - Calculus an Applied Approach

Question

Find \(d y / d u, d u / d x,\) and \(d y / d x.\) $$ y=u^{2 / 3}, u=5 x^{4}-2 x $$

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(u\) is \(\frac{2}{3} u^{-1/3}\), the derivative of \(u\) with respect to \(x\) is \(20x^3 - 2\), and the derivative of \(y\) with respect to \(x\) is \((\frac{2}{3} (5x^4 - 2x)^{-1/3}) \cdot (20x^3 - 2)\).

Step 1: Compute \(d y / d u\)

By applying the power rule to the function \(y = u^{2 / 3}\), \(d y / d u\) is obtained; the rule states that if \(y = u^n\), then \(d y / d u = n \cdot u^{n-1}\). So, in this case, \( d y / d u = \frac{2}{3} u^{2/3 - 1} = \frac{2}{3} u^{-1/3}\).

Step 2: Compute \(d u / d x\)

Next, compute the first derivative of \(u\) with respect to \(x\), \(u = 5x^4 - 2x\). The power rule is used again to get: \(d u / dx = 5 \cdot 4x^{4-1} - 2 = 20x^3 - 2\).

Step 3: Compute \(d y / d x\)

Finally, compute \(d y / d x\) by applying the chain rule. The chain rule states that \(d y / d x = d y / d u \cdot d u / d x\). Thus, the derivative \(d y / d x = (\frac{2}{3} u^{-1/3}) \cdot (20x^3 - 2)\). However because \(u = 5x^4 - 2x\), replace \(u\) in the derivative. The final result is \(d y / d x = (\frac{2}{3} (5x^4 - 2x)^{-1/3}) \cdot (20x^3 - 2) \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. In a simple form, when we have a function defined as the composition of two other functions, say \( y = f(g(x)) \), the chain rule provides a way to find the derivative of \( y \) with respect to \( x \). This is particularly useful when functions are nested within each other.

When applying the chain rule, the key idea is to break down the derivative into two manageable parts: the derivative of the outer function with respect to the inner function (\( g(x) \)), and the derivative of the inner function itself with respect to \( x \). The rule is mathematically expressed as:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

In our original problem, we applied the chain rule to find \( \frac{dy}{dx} \) by first computing \( \frac{dy}{du} \) for the function \( y = u^{2/3} \) and \( \frac{du}{dx} \) for the function \( u = 5x^4 - 2x \). This means that the accurate application of the chain rule allows for concise derivative calculation of complex, layered expressions. Differentiating composite functions like these is a staple in deeper calculus problems.

Applying the Power Rule

The power rule is an essential tool for differentiating expressions where a variable is raised to a constant power. It simplifies what might otherwise be labor-intensive calculations, making it an indispensable rule in calculus differentiation.

The power rule states: if you have a function in the form \( y = u^n \), then its derivative with respect to \( u \) is \( \frac{dy}{du} = n \cdot u^{n-1} \). This transformation comes from the general methodology of decreasing the exponent by one and multiplying by the original exponent value.

• Example: For \( y = u^{2/3} \), applying the power rule gives \( \frac{dy}{du} = \frac{2}{3} u^{-1/3} \).

This rule can also be extended to variables other than \( x \) or \( u \), making it widely adaptable. In step 2 of our exercise, the power rule was applied again to find \( \frac{du}{dx} \) for the expression \( u = 5x^4 - 2x \), yielding \( 20x^3 - 2 \). It is by applying the power rule efficiently that makes derivative calculation both straightforward and practical for intricate functions.

Complete Derivative Calculation

Derivative calculation is the process of finding the derivative of a function, which is the measure of how a function's output value changes as its input value changes. It's a core task in calculus that finds broad applications in sciences and engineering. When differentiating a composite function, combining different derivative rules is necessary.

In our example, to find \( \frac{dy}{dx} \), we followed the steps using both the chain rule and the power rule. These steps included calculating \( \frac{dy}{du} \) and \( \frac{du}{dx} \) separately. Using the chain rule, we then multiply these results to get the complete derivative:

• \( \frac{dy}{dx} = \left(\frac{2}{3} u^{-1/3}\right) \cdot \left(20x^3 - 2\right) \)
• With substitution: \( \frac{dy}{dx} = \frac{2}{3} \left(5x^4 - 2x\right)^{-1/3} \cdot \left(20x^3 - 2\right) \).

This comprehensive process exemplifies how derivatives are computed for nested functions, integrating several fundamental rules of calculus. It ensures precision and clarity, guiding you through what might initially seem as complex calculations.