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Chapter 1: Problem 1

Find \(\frac{d y}{d x}\). $$y=x^{7}$$

Short Answer

Expert verified
\( \frac{d y}{d x} = 7 x^{6} \)

Step by step solution

01

- Understand the Problem

Given a function \( y = x^{7} \), the goal is to find its derivative with respect to \( x \).
02

- Recall the Power Rule

The power rule for differentiation states that if \( y = x^n \), where \( n \) is a constant, then \( \frac{d y}{d x} = n x^{n-1} \).
03

- Apply the Power Rule

Since \( y = x^7 \) and \( n = 7 \), applying the power rule gives \( \frac{d y}{d x} = 7 x^{7-1} \).
04

- Simplify the Expression

Simplify the exponent to get \( \frac{d y}{d x} = 7 x^{6} \).

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Key Concepts

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

Power Rule
The power rule is one of the most straightforward rules in differentiation. It simplifies the process of finding the derivative for functions of the form \(y = x^n\). The power rule states: if you have \(y = x^n\), then the derivative \(\frac{dy}{dx}\) is found by bringing down the exponent as a coefficient and then reducing the exponent by one. In mathematical terms, this is written as \(\frac{d}{dx} x^n = n x^{n-1}\). This means if your function is \(y = x^7\), using the power rule, the derivative would be \(\frac{dy}{dx} = 7 x^{6}\).
Differentiation
Differentiation is a fundamental concept in calculus that deals with computing the derivative of a function. The derivative tells us how a function changes as its input changes. In simpler terms, it gives us the slope of the function at any point. For the function \(y = x^7\), differentiation involves applying the power rule to find that \(\frac{d y}{d x} = 7 x^{6}\). Derivatives are used extensively in fields like physics, engineering, and economics to model change and optimize solutions. To summarize: differentiation helps us understand the rate at which quantities change and find the slopes of curves.
Simplifying Expressions
In calculus, after applying differentiation rules like the power rule, it's crucial to simplify the resulting expression. Simplifying ensures that the derivative is in its simplest form, making it easier to understand and use in further calculations. In our example, we differentiated \(y = x^7\) and initially obtained \(7 x^{7-1}\). By simplifying the exponent \(7-1\) to \(6\), we get the clean and simplified derivative: \(7 x^{6}\). Always simplify where possible to avoid unnecessary complexity. Simplification also helps to spot errors and makes the expression easier to interpret and apply in various scenarios.

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Most popular questions from this chapter

Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$f(x)=\sqrt{x^{2}+\sqrt{1-3 x}}$$

Differentiate each function. $$f(x)=\frac{x}{x^{-1}+1}$$

Find \(d y / d x .\) Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$y=\frac{x^{5}-x^{3}}{x^{2}}$$

A proof of the Product Rule appears below. Provide a justification for each step. $$\text { a) } \frac{d}{d x}[f(x) \cdot g(x)]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x)}{h}$$ $$\text { b) } \quad=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x+h) g(x)+f(x+h) g(x)-f(x) g(x)}{h}$$ $$\text { c) } \quad=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x+h) g(x)}{h}+\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)}{h}$$ $$\begin{array}{ll} \text { d) } & =\lim _{h \rightarrow 0}\left[f(x+h) \cdot \frac{g(x+h)-g(x)}{h}\right]+\lim _{h \rightarrow 0}\left[g(x) \cdot \frac{f(x+h)-f(x)}{h}\right] \end{array}$$ $$\text { e) } \quad=f(x) \cdot \lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}+g(x) \cdot \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ $$\mathbf{f}) \quad=f(x) \cdot g^{\prime}(x)+g(x) \cdot f^{\prime}(x)$$ $$\text { g) } \quad=f(x) \cdot\left[\frac{d}{d x} g(x)\right]+g(x) \cdot\left[\frac{d}{d x} f(x)\right]$$

Differentiate. $$g(x)=\sqrt{\frac{x^{2}-4 x}{2 x+1}}$$
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