Questions: Find (y') and (y''). [ y=ln (sec (9 x)+tan (9 x)) y'=square y''=square ]

Find (y') and (y''). [ y=ln (sec (9 x)+tan (9 x)) y'=square y''=square ]
Transcript text: Find $y^{\prime}$ and $y^{\prime \prime}$. \[ \begin{array}{l} y=\ln (\sec (9 x)+\tan (9 x)) \\ y^{\prime}=\square \\ y^{\prime \prime}=\square \end{array} \]
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Solution

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Solution Steps

To find the first and second derivatives of the given function \( y = \ln (\sec (9x) + \tan (9x)) \), we will use the chain rule and the derivatives of trigonometric functions.

  1. First, find \( y' \) by differentiating \( y \) with respect to \( x \).
  2. Then, find \( y'' \) by differentiating \( y' \) with respect to \( x \).
Step 1: Define the Function

Given the function: \[ y = \ln (\sec (9x) + \tan (9x)) \]

Step 2: Compute the First Derivative

To find the first derivative \( y' \), we use the chain rule: \[ y' = \frac{d}{dx} \ln (\sec (9x) + \tan (9x)) \] \[ y' = \frac{1}{\sec (9x) + \tan (9x)} \cdot \frac{d}{dx} (\sec (9x) + \tan (9x)) \] \[ \frac{d}{dx} (\sec (9x) + \tan (9x)) = 9 \sec (9x) \tan (9x) + 9 \sec^2 (9x) \] Thus, \[ y' = \frac{9 \sec (9x) \tan (9x) + 9 \sec^2 (9x)}{\sec (9x) + \tan (9x)} \] \[ y' = \frac{9 (\sec (9x) \tan (9x) + \sec^2 (9x))}{\sec (9x) + \tan (9x)} \]

Step 3: Compute the Second Derivative

To find the second derivative \( y'' \), we differentiate \( y' \) with respect to \( x \): \[ y'' = \frac{d}{dx} \left( \frac{9 (\sec (9x) \tan (9x) + \sec^2 (9x))}{\sec (9x) + \tan (9x)} \right) \]

Using the quotient rule: \[ y'' = \frac{(\sec (9x) + \tan (9x)) \cdot \frac{d}{dx} [9 (\sec (9x) \tan (9x) + \sec^2 (9x))] - 9 (\sec (9x) \tan (9x) + \sec^2 (9x)) \cdot \frac{d}{dx} (\sec (9x) + \tan (9x))}{(\sec (9x) + \tan (9x))^2} \]

Simplifying the derivatives: \[ \frac{d}{dx} [9 (\sec (9x) \tan (9x) + \sec^2 (9x))] = 81 \sec (9x) \tan^2 (9x) + 81 \sec^3 (9x) + 81 \sec (9x) \] \[ \frac{d}{dx} (\sec (9x) + \tan (9x)) = 9 \sec (9x) \tan (9x) + 9 \sec^2 (9x) \]

Thus, \[ y'' = \frac{(9 \sec (9x) \tan (9x) + 9 \sec^2 (9x)) \cdot (\sec (9x) + \tan (9x)) - 9 (\sec (9x) \tan (9x) + \sec^2 (9x)) \cdot (9 \sec (9x) \tan (9x) + 9 \sec^2 (9x))}{(\sec (9x) + \tan (9x))^2} \]

Final Answer

The first derivative is: \[ y' = \frac{9 \tan^2 (9x) + 9 \tan (9x) \sec (9x) + 9}{\tan (9x) + \sec (9x)} \]

The second derivative is: \[ y'' = \frac{81 \sec (9x) \tan^2 (9x) + 81 \sec^3 (9x) + 81 \sec (9x)}{\tan (9x) + \sec (9x)} - \frac{(9 \tan^2 (9x) + 9 \tan (9x) \sec (9x) + 9)^2}{(\tan (9x) + \sec (9x))^2} \]

\[ \boxed{y' = \frac{9 \tan^2 (9x) + 9 \tan (9x) \sec (9x) + 9}{\tan (9x) + \sec (9x)}} \]

\[ \boxed{y'' = \frac{81 \sec (9x) \tan^2 (9x) + 81 \sec^3 (9x) + 81 \sec (9x)}{\tan (9x) + \sec (9x)} - \frac{(9 \tan^2 (9x) + 9 \tan (9x) \sec (9x) + 9)^2}{(\tan (9x) + \sec (9x))^2}} \]

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