The given expression is:

$f(x) = \frac{-36x^4 + 20x^3 - 127x^2 + 20x - 30}{(15x^3 - 6x^2 + 10x - 4)^2}$

To calculate the derivative of this rational function, we'll apply the quotient rule, which states:

$\left( \frac{u(x)}{v(x)} \right)' = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{v(x)^2}$

Where $u(x)$ is the numerator and $v(x)$ is the denominator.

Step 1: Define $u(x)$ and $v(x)$

• $u(x) = -36x^4 + 20x^3 - 127x^2 + 20x - 30$
• $v(x) = (15x^3 - 6x^2 + 10x - 4)^2$

We need to compute $u'(x)$ and $v'(x)$.

Step 2: Differentiate $u(x)$

Let's differentiate the numerator $u(x) = -36x^4 + 20x^3 - 127x^2 + 20x - 30$ term by term:

$u'(x) = \frac{d}{dx} \left(-36x^4\right) + \frac{d}{dx} \left(20x^3\right) + \frac{d}{dx} \left(-127x^2\right) + \frac{d}{dx} \left(20x\right) + \frac{d}{dx} \left(-30\right)$

• $\frac{d}{dx} (-36x^4) = -144x^3$
• $\frac{d}{dx} (20x^3) = 60x^2$
• $\frac{d}{dx} (-127x^2) = -254x$
• $\frac{d}{dx} (20x) = 20$
• $\frac{d}{dx} (-30) = 0$

So, $u'(x) = -144x^3 + 60x^2 - 254x + 20$

Step 3: Differentiate $v(x)$

We have $v(x) = (15x^3 - 6x^2 + 10x - 4)^2$. This is a composition of functions, so we'll apply the chain rule.

Let $g(x) = 15x^3 - 6x^2 + 10x - 4$. Then, $v(x) = g(x)^2$.

Using the chain rule: $v'(x) = 2g(x) \cdot g'(x)$

First, differentiate $g(x) = 15x^3 - 6x^2 + 10x - 4$:

$g'(x) = \frac{d}{dx} (15x^3) + \frac{d}{dx} (-6x^2) + \frac{d}{dx} (10x) + \frac{d}{dx} (-4)$ $g'(x) = 45x^2 - 12x + 10$

Now, substitute back into the chain rule:

$v'(x) = 2(15x^3 - 6x^2 + 10x - 4)(45x^2 - 12x + 10)$

Step 4: Apply the quotient rule

Now we substitute $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$ into the quotient rule formula:

$f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{v(x)^2}$

Substitute the expressions:

$f'(x) = \frac{(15x^3 - 6x^2 + 10x - 4)^2 \cdot (-144x^3 + 60x^2 - 254x + 20) - (-36x^4 + 20x^3 - 127x^2 + 20x - 30) \cdot 2(15x^3 - 6x^2 + 10x - 4)(45x^2 - 12x + 10)}{(15x^3 - 6x^2 + 10x - 4)^4}$

Step 5: Simplify the expression

The expression is now fully written in the quotient rule form. You can further simplify this expression, but the process involves algebraic manipulation and expanding the terms, which is lengthy.

Do you want to proceed with further simplifications or leave it as is? Feel free to ask any questions about the process so far!

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Here are 5 related questions to deepen your understanding:

1. How does the chain rule apply when differentiating composite functions?
2. What is the advantage of using the quotient rule in this derivative?
3. How can you simplify large algebraic expressions after applying the quotient rule?
4. Can you verify the derivative using alternative methods like logarithmic differentiation?
5. How do you deal with higher powers in both the numerator and denominator in derivatives?

Tip: When using the quotient rule, always compute the derivatives of both the numerator and the denominator before plugging them into the formula for smoother calculations.