If `y=tan^(-1)x^3` then find `(d^2y)/(dx^(2))`.

Answer

To find the second derivative of the function \( y = \tan^{-1}(x^3) \), we will follow these steps:

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

Using the chain rule, we differentiate \( y = \tan^{-1}(u) \) where \( u = x^3 \).

\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

Here, \( \frac{du}{dx} = 3x^2 \) and \( u^2 = (x^3)^2 = x^6 \).

Thus, we have:

\[ \frac{dy}{dx} = \frac{1}{1 + x^6} \cdot 3x^2 = \frac{3x^2}{1 + x^6} \]

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

To find the second derivative, we will differentiate \( \frac{dy}{dx} \) using the quotient rule. The quotient rule states that if \( y = \frac{f(x)}{g(x)} \), then:

\[ \frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]

In our case, let \( f(x) = 3x^2 \) and \( g(x) = 1 + x^6 \).

1. Calculate \( f'(x) \):

\[ f'(x) = 6x \]

2. Calculate \( g'(x) \):

\[ g'(x) = 6x^5 \]

Now, applying the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{(6x)(1 + x^6) - (3x^2)(6x^5)}{(1 + x^6)^2} \]

### Step 3: Simplify the expression

Now we simplify the numerator:

\[ = \frac{6x + 6x^7 - 18x^7}{(1 + x^6)^2} \]

Combine like terms:

\[ = \frac{6x - 12x^7}{(1 + x^6)^2} \]

### Final Result

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = \frac{6x - 12x^7}{(1 + x^6)^2} \]