If `y= (x^2 +3)/( x^3 + 2x)`, find `(dy)/( dx)` at `x=1`.

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \frac{x^2 + 3}{x^3 + 2x}\) at \(x = 1\), we will use the quotient rule for differentiation. The quotient rule states that if \(y = \frac{u}{v}\), then:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \(u = x^2 + 3\) and \(v = x^3 + 2x\).

### Step 1: Identify \(u\) and \(v\)
Let:
- \(u = x^2 + 3\)
- \(v = x^3 + 2x\)

### Step 2: Differentiate \(u\) and \(v\)
Now, we differentiate \(u\) and \(v\):
- \(\frac{du}{dx} = 2x\)
- \(\frac{dv}{dx} = 3x^2 + 2\)

### Step 3: Apply the Quotient Rule
Using the quotient rule:

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

### Step 4: Substitute \(x = 1\)
Now, we substitute \(x = 1\) into the expression:

1. Calculate \(u\) and \(v\) at \(x = 1\):
- \(u = 1^2 + 3 = 4\)
- \(v = 1^3 + 2(1) = 3\)

2. Calculate \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) at \(x = 1\):
- \(\frac{du}{dx} = 2(1) = 2\)
- \(\frac{dv}{dx} = 3(1^2) + 2 = 5\)

3. Substitute into the quotient rule:
\[ \frac{dy}{dx} = \frac{(3)(2) - (4)(5)}{(3)^2} \]

### Step 5: Simplify the Expression
Now, simplify the expression:
\[ \frac{dy}{dx} = \frac{6 - 20}{9} = \frac{-14}{9} \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = 1\) is:
\[ \frac{dy}{dx} = -\frac{14}{9} \]