Find the second derivatives of the following functions :
`tanx+secx`

Answer

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

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

1. Differentiate \( y = \tan x + \sec x \).
2. We know that the derivative of \( \tan x \) is \( \sec^2 x \) and the derivative of \( \sec x \) is \( \sec x \tan x \).
3. Therefore, the first derivative is:
\[ \frac{dy}{dx} = \sec^2 x + \sec x \tan x \]

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

1. Now we differentiate \( \frac{dy}{dx} = \sec^2 x + \sec x \tan x \).
2. We will differentiate each term separately.
- For the first term \( \sec^2 x \):
- Using the chain rule, the derivative is \( 2 \sec^2 x \cdot \sec x \tan x = 2 \sec^2 x \tan x \).
- For the second term \( \sec x \tan x \):
- We apply the product rule:
\[ \frac{d}{dx}(\sec x \tan x) = \sec x \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(\sec x) \]
- The derivative of \( \tan x \) is \( \sec^2 x \) and the derivative of \( \sec x \) is \( \sec x \tan x \).
- Thus, we have:
\[ \sec x \cdot \sec^2 x + \tan x \cdot \sec x \tan x = \sec^3 x + \sec x \tan^2 x \]

3. Combining both results, we get:
\[ \frac{d^2y}{dx^2} = 2 \sec^2 x \tan x + \sec^3 x + \sec x \tan^2 x \]

### Final Result

Thus, the second derivative of the function \( y = \tan x + \sec x \) is:
\[ \frac{d^2y}{dx^2} = 2 \sec^2 x \tan x + \sec^3 x + \sec x \tan^2 x \]

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