Find the second derivatives of the following functions :
`x^(3)+tanx`

Answer

To find the second derivative of the function \( f(x) = x^3 + \tan x \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)

We start by differentiating the function \( f(x) \):

\[ f(x) = x^3 + \tan x \]

Using the power rule for \( x^3 \) and the derivative of \( \tan x \):

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(\tan x) \]

The derivative of \( x^3 \) is \( 3x^2 \) and the derivative of \( \tan x \) is \( \sec^2 x \):

\[ f'(x) = 3x^2 + \sec^2 x \]

### Step 2: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \) to find the second derivative:

\[ f'(x) = 3x^2 + \sec^2 x \]

Differentiating again:

\[ f''(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(\sec^2 x) \]

The derivative of \( 3x^2 \) is \( 6x \). To differentiate \( \sec^2 x \), we use the chain rule:

\[ \frac{d}{dx}(\sec^2 x) = 2\sec^2 x \tan x \]

Thus, we have:

\[ f''(x) = 6x + 2\sec^2 x \tan x \]

### Final Result

The second derivative of the function \( f(x) = x^3 + \tan x \) is:

\[ f''(x) = 6x + 2\sec^2 x \tan x \]

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