Question Statement

Find $\frac{dy}{dx}$ for the function:

$y=\frac{3x+4}{x^2+1}$

---

Background and Explanation

This problem involves differentiating a rational function where one polynomial is divided by another. When differentiating a quotient of two functions, we must apply the quotient rule, which provides a formula for finding the derivative of $\frac{u}{v}$ based on the individual derivatives of $u$ and $v$.

---

Solution

We are given $y=\frac{3x+4}{x^2+1}$. To differentiate this quotient, we apply the quotient rule.

First, recall the quotient rule formula:

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

Let $u=3x+4$ and $v=x^2+1$. Then we find the individual derivatives:

• $\frac{du}{dx}=3$
• $\frac{dv}{dx}=2x$

Applying the quotient rule:

$\frac{dy}{dx}=\frac{(x^2+1)\frac{d}{dx}(3x+4)-(3x+4)\frac{d}{dx}(x^2+1)}{(x^2+1{)}^2}$

Substituting the derivatives:

$=\frac{(x^2+1)(3+0)-(3x+4)(2x+0)}{(x^2+1{)}^2}$

Now expand the numerator:

$=\frac{3x^2+3-6x^2-8x}{(x^2+1{)}^2}$

Combine like terms in the numerator ($3x^2-6x^2=-3x^2$):

$\frac{dy}{dx}=\frac{-3x^2-8x+3}{(x^2+1{)}^2}$

---

Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$ (used to differentiate $x^2$ and $x$ terms)
• Constant Rule: $\frac{d}{dx}(c)=0$ (used for the constant 4 and 1)

---

Summary of Steps

1. Identify components: Let $u=3x+4$ (numerator) and $v=x^2+1$ (denominator)
2. Differentiate separately: Calculate $\frac{du}{dx}=3$ and $\frac{dv}{dx}=2x$
3. Apply quotient rule: Substitute into $\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
4. Expand: Multiply out $(x^2+1)(3)$ and $(3x+4)(2x)$ in the numerator
5. Simplify: Combine like terms ($3x^2-6x^2=-3x^2$) and arrange in standard form
6. Final answer: $\frac{dy}{dx}=\frac{-3x^2-8x+3}{(x^2+1{)}^2}$