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Compute the derivative function \(f^{\prime}(x)\) using (2.1) or (2.2) $$f(x)=\frac{2}{2 x-1}$$

Short Answer

Expert verified
The derivative of the function \(f(x) = \frac{2}{2x - 1}\) is \(f^{\prime}(x) = \frac{-4}{(2x - 1)^2}\).

Step by step solution

01

Identify the functions

Here, \(g(x) = 2\) and \(h(x) = 2x - 1\). There are no \(x\)'s in function \(g(x)\), so the derivative \(g^{\prime}(x) = 0\). For \(h(x)\) it's a simple linear function, so its derivative \(h^{\prime}(x) = 2\).
02

Apply the quotient rule

Using the quotient rule, we have: \(f^{\prime}(x) = \frac{g^{\prime}(x)h(x) - g(x)h^{\prime}(x)}{[h(x)]^2}\). Substituting the derivatives of \(g\) and \(h\) and the original functions into this formula, we get: \(f^{\prime}(x) = \frac{0\cdot(2x - 1) - 2\cdot2}{(2x-1)^2}\). Simplifying inside the numerator, we get \(f^{\prime}(x) = \frac{-4}{(2x-1)^2}\).
03

Simplify the derivative

The final step is to simplify the derivative. In this case, the derivative is already in a simplest form so no further simplification is required. Thus, the derivative of \(f(x)= \frac{2}{2x-1}\) is \(f^{\prime}(x) = \frac{-4}{(2x - 1)^2}\).

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Key Concepts

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

Derivative function
A derivative function represents the rate at which a function changes as its input changes. In calculus, this is foundational for understanding how functions behave. The notation \( f^{\prime}(x) \) is used to represent the derivative of a function \( f(x) \). It tells us the slope of the tangent line to the curve \( f(x) \) at any point \( x \).
When calculating the derivative, we're essentially determining how a function's output responds to infinitesimally small changes in its input, which is vital for analyzing real-world scenarios like speed, growth, and trends.
For example, finding the derivative of a function allows us to predict changes and optimize different conditions based on current trends.
  • Derivatives help understand the dynamic behavior of changing systems.
  • They provide instantaneous rates of change.
  • Enable the solving of complex problems in physics, engineering, and economics.
Quotient rule
In calculus, the quotient rule is a method for finding the derivative of a quotient of two functions. When you have a function written as the division of two differentiable functions \( g(x) \) and \( h(x) \), the quotient rule helps compute its derivative.
The formula is: \[ f^{\prime}(x) = \frac{g^{\prime}(x) h(x) - g(x) h^{\prime}(x)}{[h(x)]^2} \]This rule is particularly useful when the function at hand is a fraction, complicating direct differentiation.
To use the quotient rule, follow these steps:
  • Differentiating \( g(x) \) to find \( g^{\prime}(x) \).
  • Differentiating \( h(x) \) to find \( h^{\prime}(x) \).
  • Substituting these into the formula above.
Remember, the correct differentiation of the numerator and denominator separately is crucial, as errors there can affect the entire solution.
Simplifying derivatives
Simplifying derivatives is a key part of the differentiation process. Once you've applied rules like the quotient rule, it's important to simplify the expression to obtain a clear, straightforward answer.
Simplification ensures that the derivative is easy to interpret and use, especially if you need to evaluate it at specific points or compare it with other functions.
Some tips for simplifying derivatives include:
  • Combine like terms to make the expression shorter.
  • Factor out common factors to make the form neater.
  • Reduce fractions when possible for clarity.
For instance, after applying the quotient rule to \( f(x) = \frac{2}{2x-1} \), the derivative \( f^{\prime}(x) = \frac{-4}{(2x - 1)^2} \) is already simplified because it can't be reduced further.
Simplification is less about procedure and more about practice, clarity, and precision, ensuring that the derivative is usable and understandable in subsequent calculations or applications.

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Most popular questions from this chapter

For different positive values of \(k,\) determine how many times \(y=\sin k x\) intersects \(y=x .\) In particular, what is the largest value of \(k\) for which there is only one intersection? Try to determine the largest value of \(k\) for which there are three intersections.

Suppose the function \(v(d)\) represents the average speed in m/s of the world record running time for \(d\) meters. For example, if the fastest 200 -meter time ever is \(19.32 \mathrm{s}\), then \(v(200)=200 / 19.32 \approx 10.35 . \quad\) Compare the function \(f(d)=26.7 d^{-0.177}\) to the values of \(v(d),\) which you will have to research and compute, for distances ranging from \(d=400\) to \(d=2000 .\) Explain what \(v^{\prime}(d)\) would represent.

The table shows the percentage of English Premier League soccer players by birth month, where \(x=0\) represents November, \(x=1\) represents December and so on. (The data are adapted from John Wesson's The Science of Soccer.) If these data come from a differentiable function \(f(x),\) estimate \(f^{\prime}(1) .\) Interpret the derivative in terms of the effect of being a month older but in the same grade of school. $$\begin{array}{|c|c|c|c|c|c|} \hline \text { Month } & 0 & 1 & 2 & 3 & 4 \\ \hline \text { Percent } & 13 & 11 & 9 & 7 & 7 \\ \hline \end{array}$$

The concentration of a certain chemical after \(t\) seconds of an autocatalytic reaction is given by \(x(t)=\frac{10}{9 e^{-10 t}+2} .\) Show that \(x^{\prime}(t)>0\) and use this information to determine that the concentration of the chemical never exceeds 5 .

For \(f(x)=\\{\begin{array}{ll}\frac{\sin x}{x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0\end{array}\) show . that \(f\) is continuous and differentiable for all \(x\). (Hint: Focus on \(x=0\) )

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