Find `(a)dy/dxand(b)(d^(2)y)/(dx^(2))` when y is given by :
`1/(2x+3),xne-3/2`

Answer

To solve the problem of finding \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) for the function \(y = \frac{1}{2x + 3}\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dx}\)

Given:
\[ y = \frac{1}{2x + 3} \]

We will use the quotient rule for differentiation. The quotient rule states that if \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

In our case:
- \(u = 1\) (constant)
- \(v = 2x + 3\)

Now, we calculate the derivatives:
- \(\frac{du}{dx} = 0\) (since the derivative of a constant is zero)
- \(\frac{dv}{dx} = 2\) (derivative of \(2x + 3\))

Now, applying the quotient rule:
\[ \frac{dy}{dx} = \frac{(2x + 3)(0) - (1)(2)}{(2x + 3)^2} \]
\[ = \frac{-2}{(2x + 3)^2} \]

### Step 2: Find \(\frac{d^2y}{dx^2}\)

Now, we need to find the second derivative \(\frac{d^2y}{dx^2}\). We will differentiate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{2}{(2x + 3)^2} \]

Using the quotient rule again:
Let \(u = -2\) and \(v = (2x + 3)^2\).

Now, we calculate the derivatives:
- \(\frac{du}{dx} = 0\)
- To find \(\frac{dv}{dx}\), we use the chain rule:
\[ \frac{dv}{dx} = 2(2x + 3) \cdot 2 = 4(2x + 3) \]

Now, applying the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(2x + 3)^2(0) - (-2)(4(2x + 3))}{((2x + 3)^2)^2} \]
\[ = \frac{8(2x + 3)}{(2x + 3)^4} \]
\[ = \frac{8}{(2x + 3)^3} \]

### Final Answers:
(a) \(\frac{dy}{dx} = -\frac{2}{(2x + 3)^2}\)

(b) \(\frac{d^2y}{dx^2} = \frac{8}{(2x + 3)^3}\)