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

Answer

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

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

Using the chain rule, we know that the derivative of \( \tan^{-1}(u) \) is given by:

\[ \frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

Here, \( u = x^3 \). Therefore, we first find \( \frac{du}{dx} \):

\[ \frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2 \]

Now, applying the chain rule:

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

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

To find the second derivative, we need to differentiate \( \frac{dy}{dx} \):

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

Using the quotient rule, which states that if \( y = \frac{u}{v} \), then:

\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \]

Let \( u = 3x^2 \) and \( v = 1 + x^6 \).

First, we find \( u' \) and \( v' \):

\[ u' = \frac{d}{dx}(3x^2) = 6x \]
\[ v' = \frac{d}{dx}(1 + x^6) = 6x^5 \]

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:

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

Expanding the numerator:

\[ = \frac{6x + 6x^7 - 18x^7}{(1 + x^6)^2} \]
\[ = \frac{6x - 12x^7}{(1 + x^6)^2} \]

Thus, the second derivative is:

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

### Final Answer

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

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