y= (5x^(5)-x^(3)-4)/(2x^(2)-5)                                                                                                                                                   The numerator of y' is

To find the numerator of \( y' \) (the derivative of \( y \)) where

\[
y = \frac{5x^5 - x^3 - 4}{2x^2 - 5}
\]

we'll need to apply the **quotient rule** for derivatives. The quotient rule states that if \( y = \frac{u(x)}{v(x)} \), then

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

where \( u(x) \) is the numerator and \( v(x) \) is the denominator. In our case:

- \( u(x) = 5x^5 - x^3 - 4 \)
- \( v(x) = 2x^2 - 5 \)

Now let's differentiate:

1. **Find \( u'(x) \):**

\[
u'(x) = \frac{d}{dx}(5x^5 - x^3 - 4) = 25x^4 - 3x^2
\]

2. **Find \( v'(x) \):**

\[
v'(x) = \frac{d}{dx}(2x^2 - 5) = 4x
\]

Now, apply the quotient rule formula for the numerator:

\[
\text{Numerator of } y' = v(x)u'(x) - u(x)v'(x)
\]

Substituting the expressions for \( u(x), u'(x), v(x), \) and \( v'(x) \):

\[
\text{Numerator of } y' = (2x^2 - 5)(25x^4 - 3x^2) - (5x^5 - x^3 - 4)(4x)
\]

This is the numerator of the derivative \( y' \). It can be expanded, but this is the expression that represents the numerator.

Would you like to simplify it further or need any additional details?

Here are some related questions to consider:
1. How do you simplify the expression for the numerator?
2. What are some common mistakes when using the quotient rule?
3. How would the process change if the denominator were more complex?
4. What is the significance of the quotient rule in calculus?
5. Can you identify situations where the quotient rule would not be applicable?

**Tip:** Always double-check your derivatives before applying the quotient rule—errors in the individual derivatives lead to incorrect final results.

y = (5x^5 - x^3 - 4) / (2x^2 - 5). The numerator of y' is?

Explore how to find the numerator of the derivative y' for the function y = (5x^5 - x^3 - 4) / (2x^2 - 5). This solution demonstrates the application of the quotient rule in calculus, with step-by-step differentiation of the numerator and denominator. Suitable for students in Grades 11-12.

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