If y ` = tan ^(-1) (sec x + tan x) , " find " (d^(2) y)/( dx^(2)) at x = (pi)/(4)`

Answer

To solve the problem, we need to find the second derivative of the function \( y = \tan^{-1}(\sec x + \tan x) \) at \( x = \frac{\pi}{4} \).

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

Using the chain rule for differentiation, we have:

\[ \frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot \frac{d}{dx}(\sec x + \tan x) \]

Now we need to differentiate \( \sec x + \tan x \):

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

Thus,

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

Now substituting this back into our expression for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{1 + (\sec x + \tan x)^2} \]

### Step 2: Evaluate \( \frac{dy}{dx} \) at \( x = \frac{\pi}{4} \)

At \( x = \frac{\pi}{4} \):

\[ \sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \]

So,

\[ \sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right) = \sqrt{2} + 1 \]

Now calculate \( (\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right))^2 \):

\[ (\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \]

Now substituting these values into \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\sqrt{2} \cdot 1 + 2}{1 + (3 + 2\sqrt{2})} = \frac{\sqrt{2} + 2}{4 + 2\sqrt{2}} \]

### Step 3: Simplify \( \frac{dy}{dx} \)

To simplify \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\sqrt{2} + 2}{4 + 2\sqrt{2}} = \frac{1}{2} \]

### Step 4: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

Since \( \frac{dy}{dx} = \frac{1}{2} \) is a constant, its derivative is:

\[ \frac{d^2y}{dx^2} = 0 \]

### Step 5: Conclusion

Thus, the value of \( \frac{d^2y}{dx^2} \) at \( x = \frac{\pi}{4} \) is:

\[ \frac{d^2y}{dx^2} = 0 \]