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Chapter 3: Problem 22

Find the values of the derivatives. $$\left.\frac{d w}{d z}\right|_{z=4} \quad \text { if } \quad w=z+\sqrt{z}$$

Short Answer

Expert verified
The derivative at \( z = 4 \) is 1.25.

Step by step solution

01

Identify the Function

First, recognize the given function: \( w = z + \sqrt{z} \). The goal is to find the derivative of this function with respect to \( z \) and then evaluate it at \( z = 4 \).
02

Differentiate the Function

Calculate the derivative of \( w \) with respect to \( z \). Differentiate each term individually: \[ \frac{d}{dz}(z) = 1 \text{ and } \frac{d}{dz}(\sqrt{z}) = \frac{1}{2\sqrt{z}} \]. Therefore, the derivative of \( w \) is \( \frac{dw}{dz} = 1 + \frac{1}{2\sqrt{z}} \).
03

Substitute and Evaluate

Substitute \( z = 4 \) into the derivative to find the value of the derivative at \( z = 4 \): \[ \frac{dw}{dz} \bigg|_{z=4} = 1 + \frac{1}{2\sqrt{4}} = 1 + \frac{1}{4} = 1.25 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation Techniques
Differentiation is a fundamental tool in calculus that allows us to find the rate of change of a function. When differentiating, we employ a variety of techniques depending on the function we are dealing with.
  • For polynomial functions like \( z \), the power rule is typically used. Here, the derivative of \( z^n \) is \( nz^{n-1} \).
  • The derivative of a constant is 0, and for linear terms like \( z \), the derivative simplifies directly to the coefficient of \( z \), which is 1.
  • For non-linear expressions such as square roots, special attention is needed, as these require techniques like the chain rule or special formula adaptations.
In our function \( w = z + \sqrt{z} \), we differentiate each component term by term. This approach simplifies problems, as each term is handled individually with the appropriate rule.
Function Evaluation
Once you have differentiated a function, evaluating it at a specific point allows you to find the exact rate of change at that point. This is what we're doing when we calculate the derivative at \( z = 4 \).
It's crucial to carefully substitute the specific value into the differentiated function. Replace \( z \) with 4 in the expression for the derivative you obtained. Double-check calculations to ensure precision, as this value directly relates to interpreting real-world situations or further mathematical calculations.
Square Root Differentiation
Differentiating square root functions can be a bit tricky due to their non-standard form. To successfully find the derivative of a square root like \( \sqrt{z} \), we use a specific rule:
  • The derivative of \( \sqrt{z} \) is \( \frac{1}{2\sqrt{z}} \). This comes by rewriting \( \sqrt{z} \) as \( z^{1/2} \), allowing us to apply the power rule.
  • When \( z^{1/2} \) is differentiated, you multiply by the power, and decrease the power by one, resulting in \( \frac{1}{2}z^{-1/2} \).
  • Finally, this expression can be rewritten as \( \frac{1}{2\sqrt{z}} \) to simplify interpretations and further math operations.
Understanding square root differentiation is crucial for finding the slope and changes in curves involving roots. This form of differentiation is widely used across various applications in mathematics and science.

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Most popular questions from this chapter

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L .\) $$ \begin{array}{l}{\text { a. With } L \text { held constant and } g \text { as the independent variable, }} \\ {\text { calculate } d T \text { and use it to answer parts (b) and (c). }} \\ {\text { b. If } g \text { increases, will } T \text { increase or decrease? Will a pendulum }} \\ {\text { clock speed up or slow down? Explain. }}\end{array} $$ $$ \begin{array}{l}{\text { c. A clock with a } 100 \text { -cm pendulum is moved from a location }} \\ {\text { where } g=980 \mathrm{cm} / \mathrm{sec}^{2} \text { to a new location. This increases the }} \\ {\text { period by } d T=0.001 \mathrm{sec} . \text { Find } d g \text { and estimate the value of }} \\\ {g \text { at the new location. }}\end{array} $$

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