If `y = tan^(-1)x`, then find `(d^(2)y)/(dx^(2))` in term of y alone.

Explanation

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

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

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)
Now we need to differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{1 + x^2}\right) \]
Using the quotient rule, where \( u = 1 \) and \( v = 1 + x^2 \):
\[ \frac{d^2y}{dx^2} = \frac{0 \cdot (1 + x^2) - 1 \cdot (2x)}{(1 + x^2)^2} = \frac{-2x}{(1 + x^2)^2} \]

### Step 3: Express \( x \) in terms of \( y \)
Since we need to express \( \frac{d^2y}{dx^2} \) in terms of \( y \), we know:
\[ x = \tan(y) \]

### Step 4: Substitute \( x \) into the second derivative
Now we substitute \( x = \tan(y) \) into our expression for \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{-2\tan(y)}{(1 + \tan^2(y))^2} \]

### Step 5: Simplify using the identity \( 1 + \tan^2(y) = \sec^2(y) \)
Using the identity \( 1 + \tan^2(y) = \sec^2(y) \), we can rewrite the expression:
\[ \frac{d^2y}{dx^2} = \frac{-2\tan(y)}{\sec^4(y)} \]

### Step 6: Rewrite in terms of sine and cosine
Recall that \( \tan(y) = \frac{\sin(y)}{\cos(y)} \) and \( \sec(y) = \frac{1}{\cos(y)} \):
\[ \frac{d^2y}{dx^2} = -2 \tan(y) \cos^4(y) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -2 \frac{\sin(y)}{\cos(y)} \cdot \cos^4(y) = -2 \sin(y) \cos^3(y) \]

### Final Result
Thus, the second derivative \( \frac{d^2y}{dx^2} \) in terms of \( y \) alone is:
\[ \frac{d^2y}{dx^2} = -2 \sin(y) \cos^3(y) \]