Math Problem Statement

Calculate derivative of minus 364 + 20 excuse minus 127 x square + 20 x minus 30 / 15 x cube minus 6 x square + 10 x minus 4 whole square

Solution

We need to compute the derivative of the following expression:

f(x)=364+20x127x2+20x30(15x36x2+10x4)2f(x) = \frac{-364 + 20x - 127x^2 + 20x - 30}{(15x^3 - 6x^2 + 10x - 4)^2}

Step 1: Simplify the numerator

The numerator simplifies to: N(x)=364+20x127x2+20x30=36430+40x127x2N(x) = -364 + 20x - 127x^2 + 20x - 30 = -364 - 30 + 40x - 127x^2 N(x)=394+40x127x2N(x) = -394 + 40x - 127x^2

Step 2: Identify the denominator

The denominator is: D(x)=(15x36x2+10x4)2D(x) = (15x^3 - 6x^2 + 10x - 4)^2

Step 3: Use the quotient rule

The quotient rule states that for a function f(x)=N(x)D(x)f(x) = \frac{N(x)}{D(x)}, the derivative is: f(x)=N(x)D(x)N(x)D(x)D(x)2f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{D(x)^2}

We need to find the derivatives of the numerator N(x)N(x) and the denominator D(x)D(x).

Step 4: Derivative of the numerator N(x)N(x)

N(x)=394+40x127x2N(x) = -394 + 40x - 127x^2 N(x)=402×127x=40254xN'(x) = 40 - 2 \times 127x = 40 - 254x

Step 5: Derivative of the denominator D(x)D(x)

Let’s first define D0(x)=15x36x2+10x4D_0(x) = 15x^3 - 6x^2 + 10x - 4, and note that D(x)=(D0(x))2D(x) = (D_0(x))^2. Using the chain rule, we get: D(x)=2D0(x)D0(x)D'(x) = 2D_0(x)D_0'(x)

Now, calculate D0(x)D_0'(x): D0(x)=15x36x2+10x4D_0(x) = 15x^3 - 6x^2 + 10x - 4 D0(x)=3×15x22×6x+10=45x212x+10D_0'(x) = 3 \times 15x^2 - 2 \times 6x + 10 = 45x^2 - 12x + 10

Thus: D(x)=2(15x36x2+10x4)(45x212x+10)D'(x) = 2(15x^3 - 6x^2 + 10x - 4)(45x^2 - 12x + 10)

Step 6: Apply the quotient rule

We now have all the components needed to apply the quotient rule: f(x)=(40254x)(15x36x2+10x4)2(394+40x127x2)×2(15x36x2+10x4)(45x212x+10)(15x36x2+10x4)4f'(x) = \frac{(40 - 254x)(15x^3 - 6x^2 + 10x - 4)^2 - (-394 + 40x - 127x^2) \times 2(15x^3 - 6x^2 + 10x - 4)(45x^2 - 12x + 10)}{(15x^3 - 6x^2 + 10x - 4)^4}

You can simplify this expression further if needed, but this represents the derivative of the given function.


Would you like me to simplify this expression further or help with any specific part of the process?

Here are some related questions to explore:

  1. How do we simplify the chain rule in more complex functions?
  2. Can we apply logarithmic differentiation to simplify derivative calculations?
  3. What are the general conditions for using the quotient rule in calculus?
  4. How does the degree of the polynomial in the denominator affect the final derivative?
  5. How do second derivatives behave in rational functions like this?

Tip: When differentiating a quotient, remember that it's often helpful to simplify the individual components (numerator and denominator) as much as possible before applying the quotient rule.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Quotient Rule
Polynomial Functions

Formulas

Quotient Rule: f'(x) = (N'(x)D(x) - N(x)D'(x)) / D(x)^2
Chain Rule: d/dx[f(g(x))] = f'(g(x))g'(x)
Derivative of a polynomial: d/dx [ax^n] = n * ax^(n-1)

Theorems

Quotient Rule
Chain Rule

Suitable Grade Level

Grades 11-12, College Calculus