Problem 14 - Calculus an Applied Approach

Question

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

Step-by-Step Solution

Answer

The derivative \(dy/du = -1/u^2\), the derivative \(du/dx = 3x^2 + 4x\) and the derivative \(dy/dx = -(3x^2 + 4x)/u^2\).

Step 1: Derive \(y\) with respect to \(u\)

Here, we have \(y = u^{-1}\) or \(y = 1/u\). It is quite straightforward to find the derivative of \(y\) with respect to \(u\). By applying derivative rules, we get \(dy/du = -u^{-2} = -1/(u^2)\).

Step 2: Derive \(u\) with respect to \(x\)

Now, let's find the derivative of \(u\) with respect to \(x\). We have \(u = x^3 + 2x^2\). Applying power rule for differentiation, the derivative \(du/dx = 3x^2 + 4x\).

Step 3: Derive \(y\) with respect to \(x\) using chain rule

Now we will use the chain rule \(dy/dx = (dy/du) * (du/dx)\) to find the derivative of \(y\) with respect to \(x\). Substitute the values from step 1 and step 2 into this equation, we get \( dy/dx = (-1/u^2) * (3x^2 + 4x) = -(3x^2 + 4x)/(u^2)\).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus used when dealing with composite functions. It lets us differentiate a function in terms of another variable. In simpler terms, if you have a function nested inside another, the chain rule helps you find the derivative of the outer function with respect to the inner one.
For example, in this exercise, we need to find \(dy/dx\). Because \(y\) is expressed in terms of \(u\), and \(u\) is expressed in terms of \(x\), the chain rule comes into play.
Here's how the chain rule works:

• If \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then \(dy/dx = (dy/du) \cdot (du/dx)\).

Each part is like a piece of a puzzle fitting together to help you find the overall derivative. It’s crucial to get each derivative step right to use the chain rule efficiently.

Applying the Power Rule

The power rule is another essential differentiation technique. It's used when you're dealing with functions that have powers of \(x\). The power rule states that if you have a function \(f(x) = x^n\), its derivative \(f'(x)\) is \(nx^{n-1}\). In short, you bring down the power and decrease the exponent by one.
In our original exercise, when finding \(du/dx\) for \(u = x^3 + 2x^2\), the power rule aids us tremendously:

• For \(x^3\), the derivative is \(3x^2\).
• For \(2x^2\), the derivative is \(4x\).

This gives us \(du/dx = 3x^2 + 4x\). By applying the power rule, we differentiate each term separately to find the complete derivative.

Grasping Implicit Differentiation

Implicit differentiation is used when it's difficult or impossible to solve for one variable in terms of the other. Although we didn't directly use it in this exercise, understanding it can still be valuable.
Implicit differentiation helps us find derivatives when variables are tangled in an equation. Instead of solving for one variable, you differentiate the entire equation with respect to \(x\), treating \(y\) as a function of \(x\).
Here's a quick overview:

• Differentiate each term as you normally would.
• Each time you differentiate a \(y\), multiply by \(dy/dx\).

This method is particularly useful for dealing with more complex equations where you cannot easily isolate one variable.