Question

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

Short answer

The derivatives are \( \frac{dy}{du} = \frac{2}{3} u^{-1/3} \), \( \frac{du}{dx} = 20x^3 - 2 \), and \( \frac{dy}{dx} = \frac{2}{3} ( 5x^4 - 2x )^{-1/3} \cdot ( 20x^3 - 2 ) \).

Step-by-step solution

Derivative \( \frac{dy}{du} \)

Take the derivative of \( y = u^{2/3} \) with respect to \( u \). To do this, apply the power rule which states that the derivative of \( u^n \) with respect to \( u \) is \( n \cdot u^{n-1}\). Thus, \( \frac{dy}{du} = \frac{2}{3} u^{2/3 - 1} = \frac{2}{3} u^{-1/3} \).

Derivative \( \frac{du}{dx} \)

Take the derivative of \( u = 5x^4 - 2x \) with respect to \( x \). Here, apply the power rule and the constant multiple rule. The derivative of \( x^n \) with respect to \( x \) is \( n \cdot x^{n-1} \) and the derivative of \( c \cdot f(x) \) is \( c \cdot f'(x) \) respectively. So, \( \frac{du}{dx} = 5 \cdot 4x^{4-1} - 2 = 20x^3 - 2 \).

Derivative \( \frac{dy}{dx} \)

To find \( \frac{dy}{dx} \), the chain rule is used. The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). By substitution using the results from steps 1 and 2, \( \frac{dy}{dx} = ( \frac{2}{3} u^{-1/3} ) \cdot ( 20x^3 - 2) = \frac{2}{3} ( 5x^4 - 2x )^{-1/3} \cdot ( 20x^3 - 2 ) \). Simplifying gives the result of \( \frac{dy}{dx} \).

Key concepts

Power Rule

The power rule is a fundamental technique in calculus for finding derivatives of functions. When you have a function of the form \(u^n\), the power rule helps by stating that the derivative is \(n \cdot u^{n-1}\). This is particularly useful because it simplifies the process of differentiation by providing a straightforward formula.
Let's explore this with a specific example. If we have \(y = u^{2/3}\), to find \(\frac{dy}{du}\), we'll apply the power rule. Here, \(n\) is \(\frac{2}{3}\), so the derivative becomes \(\frac{2}{3} \cdot u^{2/3 - 1}\).
When simplified, it results in \(\frac{2}{3} \cdot u^{-1/3}\). This rule is extremely versatile and applies not just to integer powers, but to any real number exponent. Therefore, mastering the power rule is crucial for anyone learning calculus, as it is a building block for more complex operations.

Chain Rule

The chain rule is an essential tool in calculus used for finding the derivative of composite functions. When you have a function where one variable depends on another, and that variable, in turn, depends on yet another, the chain rule helps you find the relationship between the outermost variables.
For example, to find \(\frac{dy}{dx}\) when \(y\) is a function of \(u\) and \(u\) is a function of \(x\), the chain rule states that you can multiply the inner derivative \(\frac{du}{dx}\) by the outer derivative \(\frac{dy}{du}\).
In the context of our exercise, we first found \(\frac{dy}{du} = \frac{2}{3} u^{-1/3}\) and \(\frac{du}{dx} = 20x^3 - 2\). The chain rule then allows us to combine these to find \(\frac{dy}{dx}\) by calculating \(\frac{dy}{du} \cdot \frac{du}{dx}\).
This concept is pivotal because many functions in calculus are composites, and understanding how to differentiate them efficiently is key to solving more complex problems.

Differentiation

Differentiation is the core process in calculus used to find the rate at which one quantity changes with respect to another. It's the process of finding the derivative, \/ pushing the understanding of how a function behaves based on its inputs.
When differentiating functions like \(u = 5x^4 - 2x\), we apply rules like the power rule or other applicable methods.
In this exercise, we first differentiated \(u\) with respect to \(x\) by using the power rule for each term: \(5x^4\) becomes \(20x^3\), and \(-2x\) becomes \(-2\).
This gives us \(\frac{du}{dx} = 20x^3 - 2\), showing how \(u\) varies as \(x\) changes. Differentiation is not only about calculation, but also understanding how small changes in the input affect the output. It forms the bedrock for many applications in physics, engineering, economics, and beyond, making it a vital skill in mathematical problem-solving.