https://file.aiquickdraw.com/mathbot/file/17242485800451usrjrcz.jpg

The image shows a function of the form:

\[
y = \frac{5x - 1}{2x + 1}
\]

To find the derivative \( \frac{dy}{dx} \), we can use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u(x)}{v(x)} \), then the derivative is given by:

\[
\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}
\]

For the given function:

- \( u(x) = 5x - 1 \)
- \( v(x) = 2x + 1 \)

Now, compute the derivatives of \( u(x) \) and \( v(x) \):

\[
u'(x) = 5
\]
\[
v'(x) = 2
\]

Applying the quotient rule:

\[
\frac{dy}{dx} = \frac{(2x + 1) \cdot 5 - (5x - 1) \cdot 2}{(2x + 1)^2}
\]

Now, simplify the expression:

\[
\frac{dy}{dx} = \frac{10x + 5 - (10x - 2)}{(2x + 1)^2}
\]

\[
\frac{dy}{dx} = \frac{10x + 5 - 10x + 2}{(2x + 1)^2}
\]

\[
\frac{dy}{dx} = \frac{7}{(2x + 1)^2}
\]

So, the derivative of the function \( y = \frac{5x - 1}{2x + 1} \) is:

\[
\frac{dy}{dx} = \frac{7}{(2x + 1)^2}
\]

Would you like any further details or have any questions about this process?

Here are 5 related questions:
1. How does the quotient rule differ from the product rule?
2. Can you apply the quotient rule to a function where both the numerator and denominator are polynomials of degree higher than 1?
3. What happens if \( u(x) \) or \( v(x) \) are constants in the quotient rule?
4. How would the derivative change if the numerator and denominator were switched?
5. What are the potential real-world applications of the derivative you just calculated?

**Tip:** Always double-check your work when applying the quotient rule to ensure all terms are simplified correctly.

Learn how to find the derivative of the function y = (5x - 1) / (2x + 1) using the quotient rule with a step-by-step solution. This problem covers differentiation techniques, specifically the quotient rule, and is suitable for advanced high school students or those studying calculus.

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