Find `(d^(2)y)/(dx^(2))` when
`y=tan^(-1)x`

Answer

To find the second derivative of \( y = \tan^{-1}(x) \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)
The first derivative of \( y = \tan^{-1}(x) \) is given by the formula:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)
To find the second derivative, we need to differentiate \( \frac{dy}{dx} \) with respect to \( x \). We will use the quotient rule for differentiation, which states:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]
In our case, \( u = 1 \) and \( v = 1 + x^2 \).

1. Differentiate \( u \):
\[ \frac{du}{dx} = 0 \]

2. Differentiate \( v \):
\[ \frac{dv}{dx} = 2x \]

Now applying the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(1 + x^2)(0) - (1)(2x)}{(1 + x^2)^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{-2x}{(1 + x^2)^2} \]

### Final Result
Thus, the second derivative of \( y = \tan^{-1}(x) \) is:
\[ \frac{d^2y}{dx^2} = \frac{-2x}{(1 + x^2)^2} \]

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