Math Problem Statement

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

Solution

To find the numerator of yy' (the derivative of yy) where

y=5x5x342x25y = \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=u(x)v(x)y = \frac{u(x)}{v(x)}, then

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

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

  • u(x)=5x5x34u(x) = 5x^5 - x^3 - 4
  • v(x)=2x25v(x) = 2x^2 - 5

Now let's differentiate:

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

u(x)=ddx(5x5x34)=25x43x2u'(x) = \frac{d}{dx}(5x^5 - x^3 - 4) = 25x^4 - 3x^2

  1. Find v(x)v'(x):

v(x)=ddx(2x25)=4xv'(x) = \frac{d}{dx}(2x^2 - 5) = 4x

Now, apply the quotient rule formula for the numerator:

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

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

Numerator of y=(2x25)(25x43x2)(5x5x34)(4x)\text{Numerator of } y' = (2x^2 - 5)(25x^4 - 3x^2) - (5x^5 - x^3 - 4)(4x)

This is the numerator of the derivative yy'. 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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Quotient Rule
Derivatives

Formulas

Quotient Rule: y' = (v(x)u'(x) - u(x)v'(x)) / (v(x))^2

Theorems

Quotient Rule

Suitable Grade Level

Grades 11-12