The function given in the image is:

$y = \frac{(x-1)(2x-3)}{x^2(2x+1)}$

To differentiate this function with respect to $x$, we will use the quotient rule, which states that if $y = \frac{f(x)}{g(x)}$, then:

$\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Step 1: Identify $f(x)$ and $g(x)$

For this function:

• $f(x) = (x-1)(2x-3)$
• $g(x) = x^2(2x+1)$

Step 2: Differentiate $f(x)$ and $g(x)$

1. Differentiate $f(x) = (x-1)(2x-3)$:

   • Use the product rule here: $f(x) = u \cdot v$ where $u = x-1$ and $v = 2x-3$.
   • Then $u' = 1$ and $v' = 2$.
   • So, $f'(x) = u'v + uv' = (1)(2x-3) + (x-1)(2)$.
   • Simplifying, $f'(x) = 2x - 3 + 2x - 2 = 4x - 5$.

2. Differentiate $g(x) = x^2(2x+1)$:

   • Again, use the product rule: $g(x) = u \cdot v$ where $u = x^2$ and $v = 2x+1$.
   • Then $u' = 2x$ and $v' = 2$.
   • So, $g'(x) = u'v + uv' = (2x)(2x+1) + (x^2)(2)$.
   • Expanding and simplifying, $g'(x) = 4x^2 + 2x + 2x^2 = 6x^2 + 2x$.

Step 3: Apply the Quotient Rule

Substitute $f(x), f'(x), g(x),$ and $g'(x)$ into the quotient rule:

$\frac{dy}{dx} = \frac{(4x - 5)x^2(2x+1) - (x-1)(2x-3)(6x^2 + 2x)}{[x^2(2x+1)]^2}$

You can simplify further if required, but this is the differentiated form of $y$ with respect to $x$.

Would you like me to simplify this expression further or provide additional details?

Related Questions:

1. What is the process for differentiating more complex fractions?
2. How does the product rule apply when differentiating composite functions?
3. Could we use logarithmic differentiation to simplify complex products and quotients?
4. What happens to the differentiation when any terms in the denominator are constants?
5. How do we find critical points after differentiating a function like this?

Tip:

For complex functions, simplifying the expression after differentiation helps in identifying key points like critical points and asymptotes.