Problem 50 - Calculus and its applications

Question

Find \(\frac{d y}{d u}, \frac{d u}{d x},\) and \(\frac{d y}{d x}\) $$ y=u(u+1) \text { and } u=x^{3}-2 x $$

Step-by-Step Solution

Answer

\( \frac{d y}{d u} = 2u + 1 \), \( \frac{d u}{d x} = 3x^2 - 2 \), and \( \frac{d y}{d x} = (2x^3 - 4x + 1)(3x^2 - 2) \).

Step 1: Differentiate y with respect to u

Given the function \( y = u(u+1) \). Use the product rule to differentiate:\[ \frac{d y}{d u} = \frac{d}{d u}[u(u+1)] = u'(u+1) + u \cdot (u+1)' \]Since \( u' = 1 \), we have: \[ \frac{d y}{d u} = 1 \cdot (u+1) + u \cdot 1 = u + 1 + u = 2u + 1 \]

Step 2: Differentiate u with respect to x

Given the function \( u = x^3 - 2x \). Differentiate \( u \) with respect to \( x \):\[ \frac{d u}{dx} = \frac{d}{dx}[x^3 - 2x] = 3x^2 - 2 \]

Step 3: Find dy/dx using the chain rule

To find \( \frac{d y}{d x} \), use the chain rule which states:\[ \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \]From previous steps, \( \frac{d y}{d u} = 2u + 1 \) and \( \frac{d u}{dx} = 3x^2 - 2 \). Therefore:\[ \frac{d y}{d x} = (2u + 1)(3x^2 - 2) \]Substitute \( u = x^3 - 2x \) into the equation:\[ \frac{d y}{d x} = (2(x^3 - 2x) + 1)(3x^2 - 2) \]Simplify:\[ \frac{d y}{d x} = (2x^3 - 4x + 1)(3x^2 - 2) \]

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are products of two simpler functions. When you have a function expressed as the product of two components, like in this exercise with \( y = u(u+1) \), understanding how to apply the product rule is essential. Here's how the product rule works:

• If you have a function \( y = f(u) \cdot g(u) \), the derivative of \( y \) with respect to \( u \) is calculated as
• \( \frac{d y}{d u} = f'(u) \cdot g(u) + f(u) \cdot g'(u) \)

Essentially, it involves differentiating each part while treating the other part as a constant, taking a sum of the products of each respective pair of functions. In this exercise, we set \( f(u) = u \) and \( g(u) = u + 1 \). Differentiating using the product rule, we find:

• \( f'(u) = 1 \)
• \( g'(u) = 1 \)

Therefore, \( \frac{d y}{d u} = 1 \cdot (u+1) + u \cdot 1 = 2u + 1 \). This result is crucial for further calculations involving chain rule.

Differentiation

Differentiation is the process of finding the derivative of a function. In simple terms, it tells us how a function changes as its input changes. For the function \( u = x^3 - 2x \) in this exercise, our goal is to find how \( u \) changes as \( x \) changes, which gives us the derivative \( \frac{d u}{d x} \).To perform differentiation here:

• Differentiate \( x^3 \), which results in \( 3x^2 \).
• Differentiate \( -2x \), which gives \( -2 \).

Combining these results, we find \( \frac{d u}{d x} = 3x^2 - 2 \).Applying differentiation correctly is essential for understanding the behavior of polynomial functions like \( u= x^3 - 2x \). It helps in calculus by determining rates of change, slopes of curves, and is foundational for more advanced topics like the chain rule.

Composite Functions

Composite functions involve the combination of two or more functions to form a new function. In calculus, when dealing with composite functions, you often need to apply the chain rule for differentiation. The function \( y = u(u+1) \) with \( u = x^3 - 2x \) is a composite function because \( y \) inherently depends on \( x \) through \( u \).To find \( \frac{d y}{d x} \), we apply the chain rule, which in essence, links the rate of change of \( y \) with respect to \( x \) via the derivative of \( u \) with respect to \( x \). Here's the simple formula:

• \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \)

With given derivatives \( \frac{d y}{d u} = 2u + 1 \) and \( \frac{d u}{d x} = 3x^2 - 2 \), the chain rule yields \( \frac{d y}{d x} = (2u + 1)(3x^2 - 2) \).Substituting \( u = x^3 - 2x \) makes it clear how the derivative depends on \( x \) alone:

• \( \frac{d y}{d x} = (2(x^3 - 2x) + 1)(3x^2 - 2) \)

Understanding composite functions is vital for navigating complex relationships in calculus.