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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 value of the derivative at \(z = 4\) is \(\frac{5}{4}\).

Step by step solution

01

Differentiate with respect to z

We need to find the derivative of the function \(w = z + \sqrt{z}\). First, differentiate each part of the function with respect to \(z\). The derivative of \(z\) with respect to \(z\) is 1. To differentiate \(\sqrt{z}\), rewrite it as \(z^{1/2}\) and use the power rule. The derivative is \((1/2)z^{-1/2}\). Thus, the derivative of \(w\) with respect to \(z\) is \(\frac{dw}{dz} = 1 + \frac{1}{2}z^{-1/2}\).
02

Simplify the derivative

The expression of the derivative is \(\frac{dw}{dz} = 1 + \frac{1}{2}z^{-1/2}\). Simplify it as much as possible. The original expression can be rewritten as \(1 + \frac{1}{2\sqrt{z}}\).
03

Substitute z = 4

Once the derivative is simplified, substitute \(z = 4\) to find the value of the derivative at that specific point. Substitute \(z = 4\) into the expression \(1 + \frac{1}{2\sqrt{z}}\), which gives \(1 + \frac{1}{2\sqrt{4}}\). Since \(\sqrt{4} = 2\), the expression simplifies to \(1 + \frac{1}{4}\).
04

Calculate the final value

Add the constants in the expression from the previous step: \(1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{5}{4}\). Thus, the derivative of \(w\) with respect to \(z\) at \(z = 4\) is \(\frac{5}{4}\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It involves analyzing how a function changes as its input changes. The process of differentiation provides the derivative, an expression that represents the slope or rate of change of a function at any given point.

Understanding differentiation requires recognizing it as the transition from one state of a variable function to another. Essentially, it helps us understand how quantities vary concerning one another.

When differentiating, several rules simplify the process. These include basic rules and more specialized ones like the chain rule or product rule. Each of these aids in addressing different types of functions without complicated calculations.

In our original exercise, we differentiated the function \( w = z + \sqrt{z} \). This involves individually differentiating each part of the function. The basic derivative rules, such as the derivative of \( z \) being 1, come into play to simplify the process.
Calculus
Calculus is the mathematical study of continuous change. It's divided mainly into two branches: differential calculus and integral calculus. Differentiation, our focus here, is part of differential calculus, which deals with the concept of the derivative.

Calculus enables us to calculate derivatives for functions, making it an essential tool in various fields such as physics, engineering, economics, and beyond. It helps us solve real-world problems by modeling and understanding complex systems.

In calculus, we learn about limits, derivatives, and the behavior of functions as they approach certain values. The derivative, like the one we calculated in the exercise, tells us the rate at which a function's output changes concerning its input.

Calculus isn't just about solving problems like finding derivatives. It also paves the way for problem-solving across probabilities, geometries, and even scenarios dealing with maximum or minimum values of functions.
Power Rule
The power rule is one of the most straightforward tools in calculus for finding derivatives of functions. It states that for a function in the form \( f(x) = x^n \), the derivative \( f'(x)\) is \( nx^{n-1} \). This rule simplifies finding derivatives of polynomial functions greatly.

When using the power rule, it's crucial to understand how exponents can be adjusted. For instance, the square root function can be rewritten as a fractional exponent, like \( \sqrt{z} = z^{1/2} \). This makes it easy to apply the power rule by bringing down the exponent and reducing it by one.

In our example with \( w = z + \sqrt{z} \), differentiating \( \sqrt{z} \) involved recognizing it as \( z^{1/2} \). Applying the power rule gave us \( (1/2)z^{-1/2} \), using the exponent relationship. It's essential for solving derivatives quickly without expanding processes unnecessarily.

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Most popular questions from this chapter

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=1-\frac{1}{u}, \quad u=g(x)=\frac{1}{1-x}, \quad x=-1$$

A light shines from the top of a pole \(15 \mathrm{m}\) high. A ball is dropped from the same height from a point \(9 \mathrm{m}\) away from the light. (See accompanying figure.) How fast is the shadow of the ball moving along the ground \(1 / 2\) s later? (Assume the ball falls a distance \(s=4.9 t^{2} \mathrm{m}\) in \(t\) seconds.)

An explosion at an oil rig located in gulf waters causes an elliptical oil slick to spread on the surface from the rig. The slick is a constant \(20 \mathrm{cm}\) thick. After several days, when the major axis of the slick is \(2 \mathrm{km}\) long and the minor axis is \(3 / 4 \mathrm{km}\) wide, it is determined that its length is increasing at the rate of \(9 \mathrm{m} / \mathrm{h}\), and its width is increasing at the rate of \(3 \mathrm{m} / \mathrm{h}\). At what rate (in cubic meters per hour) is oil flowing from the site of the rig at that time?

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right)$$

Derive the formula for the derivative with respect to \(x\) of a. sec \(x\) b. csc \(x\) c. cot \(x\)

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