Chapter 3: Problem 6
Find \(d y / d x.\) $$y=\sqrt{2} x+(1 / \sqrt{2})$$
Short Answer
Expert verified
The derivative \( \frac{dy}{dx} \) is \( \sqrt{2} \).
Step by step solution
01
Identify the Function
The given function is \( y = \sqrt{2}x + \frac{1}{\sqrt{2}} \). Our task is to find the derivative \( \frac{dy}{dx} \).
02
Identify Derivative Rules
The function \( y = \sqrt{2}x + \frac{1}{\sqrt{2}} \) is a linear function. The derivative of a linear function \( ax + b \) is simply \( a \). In this case, we will apply the derivative rule for a variable term \( ax \) and a constant term \( b \).
03
Differentiate Each Term Separately
Differentiate the term \( \sqrt{2}x \) using the power rule: the derivative of \( ax \) with respect to \( x \) is \( a \). The term \( \sqrt{2} \) is a constant factor, so \( \frac{d}{dx}(\sqrt{2}x) = \sqrt{2} \).
04
Differentiate the Constant Term
The derivative of a constant \( \frac{1}{\sqrt{2}} \) with respect to \( x \) is 0.
05
Sum of Derivatives
Combine the results from Steps 3 and 4 to find the overall derivative. Adding these, \( \frac{dy}{dx} = \sqrt{2} + 0 = \sqrt{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Functions
Linear functions are functions of the form \( y = ax + b \), where \( a \) and \( b \) are constants. These functions graph as straight lines on the Cartesian plane. The slope of the line is represented by \( a \), and \( b \) represents the y-intercept, which is where the line crosses the y-axis.
- For example, in the equation \( y = \sqrt{2}x + \frac{1}{\sqrt{2}} \), the slope \( a \) is \( \sqrt{2} \) and the y-intercept \( b \) is \( \frac{1}{\sqrt{2}} \).
- Understanding the components of a linear function aids in easily finding its derivative.
Power Rule
The power rule is a basic and essential principle in calculus. It is used to differentiate functions of the form \( x^n \), where \( n \) is any real number. The rule states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \).
- For example, if the expression is \( x^1 \), the derivative is \( 1 \cdot x^{1-1} = 1 \).
- This is why the derivative of a linear term like \( \sqrt{2}x \) simplifies to \( \sqrt{2} \), because the exponent of \( x \) is 1.
Derivative of a Constant
In calculus, the derivative of a constant term is always zero. This is because constants do not change, and differentiation measures how a function's values change as its input changes.
- For example, when evaluating \( \frac{d}{dx}\left(\frac{1}{\sqrt{2}}\right) \), we determine this term's derivative to be 0.
- Whenever a function includes a fixed number without a variable, its derivative will consistently add nothing to the result.