Problem 229 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for each function. $$y=(5-2 x)^{-2}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 4(5-2x)^{-3} \)

Step 1: Identify the Rule of Differentiation

Notice that the function given is of the form \(y=(u)^{-n}\), where \(u=5-2x\) and \(n=2\). This suggests the use of the chain rule and the power rule for differentiation.

Step 2: Apply the Power Rule

The power rule states that if \(y = u^n\), then \(\frac{d}{dx}u^n = n \cdot u^{n-1} \cdot \frac{du}{dx}\). Here, \(u = 5 - 2x\) and \(n = -2\). Thus, \(\frac{d}{dx}(5-2x)^{-2} = -2 \cdot (5-2x)^{-3} \cdot \frac{d}{dx}(5-2x)\).

Step 3: Differentiate the Inner Function

Now, find the derivative of the inner function \(u = 5 - 2x\). The derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx} = -2\).

Step 4: Substitute and Simplify

Substitute \(\frac{du}{dx} = -2\) back into the expression. So, \(\frac{d}{dx}(5-2x)^{-2} = -2 \cdot (5-2x)^{-3} \cdot (-2)\). Simplify this to get \(\frac{d y}{d x} = 4(5-2x)^{-3}\).

Key Concepts

chain rule

The chain rule is a fundamental concept in calculus used for finding the derivative of a composite function. In simple terms, a composite function is like a function within another function, such as \( y = (5 - 2x)^{-2} \). The chain rule helps us differentiate such functions by treating the outer function and the inner function separately.

When you apply the chain rule, you first find the derivative of the outer function while keeping the inner function unchanged. Then, you multiply this result by the derivative of the inner function itself. This process allows us to effectively "chain" the two functions together to get the overall derivative. It is especially useful for tackling complex expressions, making our life a lot easier in calculus!

In our given exercise, the chain rule is combined with another useful rule called the power rule, due to the specific form of our function.

power rule

The power rule is a quick and handy differentiation rule which states that if you have a function \( y = x^n \), the derivative \( \frac{d}{dx} x^n \) is simply \( nx^{n-1} \). This rule helps you take the variable's exponent, bring it down as a multiplier, and then decrease the exponent by one.

• Start with identifying the power
• Differentiate using the exponent
• Simplify the expression

This comes in handy when applying it to the inner function like in our exercise, where the entire expression \( (5-2x)^{-2} \) is raised to a power. Using the power rule, we first apply it to the expression \( (5 - 2x) \) which is the base raised to the power of \(-2\), noting that the power becomes \(-3\) post differentiation.

differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function changes. It's like looking at how steep or flat a curve is at any given point and is a core part of calculus.

The derivative can tell you numerous things, such as:

• The slope of a tangent line at a point on a curve.
• How fast or slow a quantity is changing at any moment.

To differentiate effectively, different rules are applied depending on the function's form, like the chain rule and power rule, as shown in the original solution. Combining these rules helps us navigate complex functions like the one in the exercise, allowing us to find the derivative \( \frac{dy}{dx} \) with accuracy.

inner function

In the context of the chain rule, the inner function is a crucial component that requires its own differentiation. It's essentially the "inside" part of a composite function, such as the \( u = 5 - 2x \) in our problem.

To find the derivative of a composite function, you need to focus on the inner function's own derivative. This step is vital because it directly impacts the overall derivative of the initial function when applied through the chain rule. In the exercise, differentiating the inner function \( u = 5 - 2x \) gives us \( \frac{du}{dx} = -2 \). This is then multiplied back into the derivative of the outer function, completing the differentiation process.