Chapter 3: Problem 26
Find \(d y /\left.d x\right|_{x=1}.\) $$y=\frac{4 x+1}{x^{2}-5}$$
Short Answer
Expert verified
The value of \(\left.\frac{dy}{dx}\right|_{x=1}\) is \(-\frac{13}{8}\).
Step by step solution
01
Find the Derivative of the Numerator
The numerator of the function is given as \(4x+1\). The derivative of this with respect to \(x\) is \(\frac{d}{dx}(4x + 1) = 4\).
02
Find the Derivative of the Denominator
The denominator is \(x^2 - 5\). The derivative of this with respect to \(x\) is \(\frac{d}{dx}(x^2 - 5) = 2x\).
03
Use the Quotient Rule
The function \(y = \frac{4x+1}{x^2 - 5}\) requires the use of the quotient rule which is \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot \frac{du}{dx} - u\cdot \frac{dv}{dx}}{v^2}\). Here, \(u = 4x + 1\) and \(v = x^2 - 5\).
04
Apply the Quotient Rule
Substitute the derivatives into the quotient rule: \[\frac{d}{dx}\left(\frac{4x+1}{x^2 - 5}\right) = \frac{(x^2 - 5)\cdot 4 - (4x+1)\cdot 2x}{(x^2 - 5)^2}\] This simplifies to: \[\frac{4x^2 - 20 - (8x^2 + 2x)}{(x^2 - 5)^2}\]
05
Simplify the Expression
Simplify the expression obtained in the previous step: \[\frac{4x^2 - 8x^2 - 2x - 20}{(x^2 - 5)^2} = \frac{-4x^2 - 2x - 20}{(x^2 - 5)^2}\].
06
Evaluate the Derivative at \(x = 1\)
Substitute \(x = 1\) into the derivative: \[\frac{-4(1)^2 - 2(1) - 20}{((1)^2 - 5)^2} = \frac{-4 - 2 - 20}{(1 - 5)^2}\]. Simplifying, we get \[\frac{-26}{16} = -\frac{13}{8}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Finding Derivatives
Understanding how to find the derivative of a function is a fundamental concept in calculus. Derivatives represent the rate of change of a function with respect to a variable, which is often time or space. Conceptually, finding a derivative can be thought of as finding the slope of the tangent line to the function curve at any point.To find the derivative of a function, follow these general steps:
- Identify the function you want to differentiate.
- Apply differentiation rules, such as the power rule, product rule, or quotient rule, depending on the form of your function.
- Simplify the result, if necessary.
- Evaluate the derivative at a specific point if required, as was the case when evaluating at \(x = 1\).
Rational Functions
Rational functions are expressions composed of ratios of polynomials. These can be written in the form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)eq 0\). Rational functions appear frequently in calculus and have unique behaviors due to their denominator.Key characteristics of rational functions:
- Vertical Asymptotes: These occur at values of \(x\) that make the denominator \(Q(x)\) zero, unless cancelled by the numerator.
- Horizontal Asymptotes: These describe the end behavior of the function as \(x\) approaches infinity or negative infinity.
- Intercepts: Found where the function crosses the axes, with the x-intercept obtained by setting the numerator equal to zero, and the y-intercept found by evaluating the function at \(x = 0\) if possible.
Calculus Differentiation
Calculus differentiation involves techniques to compute derivatives, an essential part of calculus that enables the understanding of changes within a system. It deals with various rules that simplify finding derivatives across different kinds of functions.Some important rules of differentiation include:
- Power Rule: For any function \(x^n\), the derivative is \(nx^{n-1}\).
- Product Rule: For two functions multiplied together, \(f(x)g(x)\), the derivative is \(f'(x)g(x) + f(x)g'(x)\).
- Quotient Rule: For a ratio of two functions \(\frac{u(x)}{v(x)}\), as seen in the exercise, the derivative is given by \( \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2} \).
- Chain Rule: Useful for finding the derivative of a composite function, where differentiating the outside function and then multiplying it by the derivative of the inside function is necessary.