differentiate with respect to x 3x+2/ 2x+3

To differentiate the expression \( \frac{3x + 2}{2x + 3} \) with respect to \(x\), we can apply the **quotient rule**. The quotient rule states that for a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative is:

\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}
\]

For \( u(x) = 3x + 2 \) and \( v(x) = 2x + 3 \):

1. Derivative of \( u(x) = 3x + 2 \):  
   \[
   u'(x) = 3
   \]
2. Derivative of \( v(x) = 2x + 3 \):  
   \[
   v'(x) = 2
   \]

Now apply the quotient rule:

\[
\frac{d}{dx} \left( \frac{3x + 2}{2x + 3} \right) = \frac{(2x + 3)(3) - (3x + 2)(2)}{(2x + 3)^2}
\]

Simplify the numerator:

\[
= \frac{6(2x + 3) - 2(3x + 2)}{(2x + 3)^2}
\]
\[
= \frac{12x + 18 - (6x + 4)}{(2x + 3)^2}
\]
\[
= \frac{12x + 18 - 6x - 4}{(2x + 3)^2}
\]
\[
= \frac{6x + 14}{(2x + 3)^2}
\]

So, the derivative is:

\[
\frac{d}{dx} \left( \frac{3x + 2}{2x + 3} \right) = \frac{6x + 14}{(2x + 3)^2}
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you apply the product rule in differentiation?
2. Can you explain the chain rule for composite functions?
3. What happens if you differentiate implicit functions?
4. How do you solve second-order derivatives for rational functions?
5. How do you apply the quotient rule when both numerator and denominator are complex functions?

**Tip**: Always simplify the numerator before dividing, it makes the derivative much easier to compute!

Discover how to differentiate the expression (3x + 2) / (2x + 3) using the quotient rule. This calculus problem is suitable for high school students in grades 11-12 or early college, focusing on the application of differentiation techniques, specifically the quotient rule.

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