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Find \(\frac{d y}{d x}\). $$y=x^{5 / 4}$$

Short Answer

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Question: Find the derivative of the function y = x^(5/4) with respect to x. Answer: The derivative of the function y = x^(5/4) with respect to x is $$\frac{dy}{dx} = \frac{5}{4}x^{\frac{1}{4}}$$.

Step by step solution

01

Identify the given function

The given function is: $$y = x^{5/4}$$
02

Apply the Power Rule to find the derivative

Using the Power Rule, the derivative of x^n is nx^(n-1). In our case, n = 5/4, thus: $$\frac{dy}{dx} = \frac{5}{4}x^{\frac{5}{4} - 1}$$
03

Simplify the exponent

Now, we need to simplify the exponent: $$\frac{5}{4} - 1 = \frac{5}{4} - \frac{4}{4} = \frac{1}{4}$$
04

Write the final derivative

Putting the simplified exponent back into our derivative, we get the final answer: $$\frac{dy}{dx} = \frac{5}{4}x^{\frac{1}{4}}$$

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Key Concepts

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

Understanding the Power Rule
When dealing with derivatives in calculus, one cannot overlook the utility of the power rule. This principle is one of the first rules you’ll learn when stepping into the realm of differentiation. The power rule states that if you have a function of the form \( y = x^n \), then the derivative, denoted as \( \frac{dy}{dx} \), is given by the formula \( nx^{n-1} \). Here, "n" represents any real number which is the exponent of \( x \).

This rule significantly simplifies the process of differentiation when working with polynomials or any other expressions that present terms like \( x^n \). By utilizing the power rule, you can quickly determine the rate of change of the function without having to revert to first principles every time.

For example, in the function \( y = x^{5/4} \), applying the power rule involves multiplying the exponent \( 5/4 \) by \( x \) raised to the new exponent which is one less than \( 5/4 \). Thus, you end up with the derivative as \( \frac{5}{4}x^{1/4} \). Remembering this simple technique helps keep the process manageable and quick.
Differentiation Basics
Differentiation is one of the two main operations in calculus, the other being integration. It involves finding the derivative of a function, which essentially tells you how a function changes at any point. In other words, it gives the slope of the tangent line to the function at a particular point.

When we differentiate a function, we are looking to find the rate of change or the velocity of a variable with respect to another. It’s like taking a snapshot of how fast something is moving, by looking at the rate of change of the position over time. Differentiation gives us insights into the behavior of functions, guiding us to know where they are increasing, decreasing, or have a relative maximum or minimum point.
  • Getting started: Always identify the function you need to differentiate first.
  • Apply known rules or techniques like the power rule, product rule, quotient rule, or chain rule.
  • Simplify your final expression to achieve the cleanest form of the derivative.
This operation is foundational to calculus and frequently used in physics, engineering, economics, and beyond to solve real-world problems.
A Dive into Calculus
Calculus is a branch of mathematics that deals with continuous change. Developed from concepts of limits, calculus allows us to deal with variables that change continuously as opposed to algebra, where variables remain constant.

Within calculus, we primarily focus on two operations: differentiation and integration. Differentiation, as discussed, is about finding the rate of change, while integration deals with finding the area under curves or combining small pieces together to see the whole picture.

In our specific problem, understanding calculus helps us comprehend why derivative formulas like the power rule work and how they apply to functions like \( y = x^{5/4} \). Through calculus, we can not only compute the slope at a point but also anticipate how the function behaves as a whole.
  • Limits give the foundation to both derivatives and integrals, ensuring solutions are precise.
  • Differential calculus focuses on rates of change, helping solve problems involving motion and growth.
  • Integral calculus, on the other hand, helps calculate quantities like areas, volumes, and total accumulated values.
In essence, calculus is all about understanding change and helps in explaining the dynamic world around us, making it indispensable in scientific exploration and various fields.

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Most popular questions from this chapter

A woman attached to a bungee cord jumps from a bridge that is \(30 \mathrm{m}\) above a river. Her height in meters above the river \(t\) seconds after the jump is \(y(t)=15\left(1+e^{-t} \cos t\right),\) for \(t \geq 0\). a. Determine her velocity at \(t=1\) and \(t=3\). b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 s. c. Use a graphing utility to estimate the maximum upward velocity.

The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by $$V(t)=\left\\{\begin{array}{ll}\frac{4}{5} t^{2} & \text { if } 0 \leq t<45 \\\\-\frac{4}{5}\left(t^{2}-180 t+4050\right) & \text { if } 45 \leq t<90, \end{array}\right.$$ where \(V\) is measured in cubic feet and \(t\) is measured in days, with \(t=0\) corresponding to May 1. a. Graph the volume function. b. Find the flow rate function \(V^{\prime}(t)\) and graph it. What are the units of the flow rate? c. Describe the flow of the stream over the 3 -month period. Specifically, when is the flow rate a maximum?

Calculate the derivative of the following functions (i) using the fact that \(b^{x}=e^{x \ln b}\) and (ii) by using logarithmic differentiation. Verify that both answers are the same. $$y=\left(x^{2}+1\right)^{x}$$

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \sqrt{10 x}$$,

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x^{2}(y-2)-e^{y}=0$$

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