To find the derivative of the given function using the quotient rule, let's first recall the quotient rule formula for a function $y = \frac{u(x)}{v(x)}$:

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

Here:

• $u(x) = 4x - 3$
• $v(x) = 5x + 4$

Step 1: Find the derivatives of $u(x)$ and $v(x)$

• $u'(x) = \frac{d}{dx}(4x - 3) = 4$
• $v'(x) = \frac{d}{dx}(5x + 4) = 5$

Step 2: Apply the quotient rule formula

$\frac{dy}{dx} = \frac{(5x + 4)(4) - (4x - 3)(5)}{(5x + 4)^2}$

Step 3: Simplify the expression

Expand both terms in the numerator:

$(5x + 4)(4) = 20x + 16$ $(4x - 3)(5) = 20x - 15$

Now subtract the two terms:

$(20x + 16) - (20x - 15) = 20x + 16 - 20x + 15 = 31$

Thus, the derivative becomes:

$\frac{dy}{dx} = \frac{31}{(5x + 4)^2}$

Final Answer:

$\frac{dy}{dx} = \frac{31}{(5x + 4)^2}$

Would you like further clarification or additional details?

Here are 5 related questions to expand on this concept:

1. What happens if we differentiate the function without the quotient rule?
2. How does the quotient rule differ from the product rule in terms of usage?
3. Can we generalize this result for any linear functions in the numerator and denominator?
4. How would the quotient rule apply if both $u(x)$ and $v(x)$ were quadratic functions?
5. What is the geometric interpretation of the quotient rule?

Tip: Always simplify the expressions for the derivative step by step to avoid errors in calculations!