Problem 12 - Thomas Calculus

Question

In Exercises \(7-12,\) find the indicated derivatives. $$ \frac{d z}{d w} \quad \text { if } \quad z=\frac{1}{\sqrt{3 w-2}} $$

Step-by-Step Solution

Answer

\( \frac{dz}{dw} = -\frac{3}{2}(3w - 2)^{-3/2} \).

Step 1: Identify the function and its components

The given function is \( z = \frac{1}{\sqrt{3w - 2}} \). This can be rewritten using a power of a negative one-half as \( z = (3w - 2)^{-1/2} \). Here, \( z \) is in terms of \( w \).

Step 2: Apply the power rule

Using the power rule for derivatives \( \frac{d}{dx}[x^n] = nx^{n-1} \), apply it to \( z = (3w - 2)^{-1/2} \). The outer function is \( u^{-1/2} \), where \( u = 3w - 2 \). Differentiate \( u^{-1/2} \) which results in \( -\frac{1}{2} u^{-3/2} \).

Step 3: Differentiate the inner function

Determine the derivative of the inner function \( u = 3w - 2 \). The derivative is \( \frac{d}{dw}(3w - 2) = 3 \).

Step 4: Use the chain rule

Apply the chain rule: \( \frac{d}{dw} z = \frac{d}{du}(u^{-1/2}) \times \frac{du}{dw} \). Substituting in our derivatives, we have:\[ \frac{dz}{dw} = -\frac{1}{2}(3w - 2)^{-3/2} \times 3 \].

Step 5: Simplify the expression

Simplify the expression from Step 4:\[ \frac{dz}{dw} = -\frac{3}{2}(3w - 2)^{-3/2} \].

Step 6: Conclusion: Final expression for the derivative

The final expression for \( \frac{dz}{dw} \) is:\[ \frac{dz}{dw} = -\frac{3}{2}(3w - 2)^{-3/2} \].

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of a composition of functions. If you have a function composed of two others, for example, a function nested within another, the chain rule helps you differentiate it efficiently.
Let's consider the example given, where you have a function in the form of

• outer function: \( (3w - 2)^{-1/2} \)
• inner function: \( u = 3w - 2 \).

When applying the chain rule, start with the outer function, differentiate it, then multiply by the derivative of the inner function.
This technique allows you to manage and tackle more complex derivative problems neatly. Make sure to keep track of each part: the first derivative affects the original power, and the second handles the inner workings, involving rates of change.

Power Rule

The power rule simplifies the process of finding derivatives for functions with powers. It's especially handy when expressions involve exponents. If you have a function of the form \( x^n \), the power rule states: \( \frac{d}{dx}[x^n] = nx^{n-1} \).
This rule is the cornerstone for differentiation in calculus. In our example, we rewrote the original function \( z = \frac{1}{\sqrt{3w - 2}} \) as \( z = (3w - 2)^{-1/2} \), setting it up perfectly to use this rule.
When applying it, remember:

• The exponent becomes a coefficient (here, \( -1/2 \)).
• Subtract one from the original exponent for the new power (hence, \( -3/2 \)).

This leaves you with a new expression which is just a hop away from the finalized derivative after applying any additional calculus rules needed.

Calculus Concepts

Calculus is fundamentally about understanding change, and derivatives are a key part of this. They represent how functions behave as inputs are altered. With derivatives, we observe the rate of change and estimated shifts for comprehensive functions.
In our exercise, multiple calculus concepts align:

• Function Composition: Seeing a function molded within another.
• Derivative Calculation: Using rules like chain and power to break down complex expressions.
• Simplification: Molding the derivative to a simpler or more useful form for further analytical or graphical work.

Combining these principles, you see the derivative not simply as a mechanical output, but as an insight into the general behavior of functions, which is both an algorithmic and conceptual achievement. This foundation is crucial for tackling advanced calculus and real-world mathematical modeling.