If `y=2 sin^(-1) (cos x)+5"cosec "^(-1) "(sec x). Find"(dy)/(dx)`.

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

\[ y = 2 \sin^{-1}(\cos x) + 5 \cos^{-1}(\sec x), \]

we will differentiate each term step by step.

### Step 1: Differentiate the first term \( 2 \sin^{-1}(\cos x) \)

Using the chain rule, the derivative of \( \sin^{-1}(u) \) is given by

\[ \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \]

Here, \( u = \cos x \), so

\[ \frac{du}{dx} = -\sin x. \]

Thus, the derivative of the first term is:

\[ \frac{d}{dx} [2 \sin^{-1}(\cos x)] = 2 \cdot \frac{1}{\sqrt{1 - \cos^2 x}} \cdot (-\sin x). \]

Since \( 1 - \cos^2 x = \sin^2 x \), we have:

\[ \sqrt{1 - \cos^2 x} = \sin x. \]

Therefore, the derivative simplifies to:

\[ \frac{d}{dx} [2 \sin^{-1}(\cos x)] = 2 \cdot \frac{-\sin x}{\sin x} = -2. \]

### Step 2: Differentiate the second term \( 5 \cos^{-1}(\sec x) \)

Using the chain rule again, the derivative of \( \cos^{-1}(u) \) is given by

\[ \frac{d}{dx} \cos^{-1}(u) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \]

Here, \( u = \sec x \), so

\[ \frac{du}{dx} = \sec x \tan x. \]

Thus, the derivative of the second term is:

\[ \frac{d}{dx} [5 \cos^{-1}(\sec x)] = 5 \cdot \left(-\frac{1}{\sqrt{1 - \sec^2 x}} \cdot \sec x \tan x\right). \]

Since \( \sec^2 x - 1 = \tan^2 x \), we have:

\[ 1 - \sec^2 x = -\tan^2 x \implies \sqrt{1 - \sec^2 x} = i \tan x. \]

Thus, the derivative simplifies to:

\[ \frac{d}{dx} [5 \cos^{-1}(\sec x)] = -5 \cdot \frac{\sec x \tan x}{i \tan x} = -5 \sec x. \]

### Step 3: Combine the derivatives

Now, we combine the derivatives from both terms:

\[ \frac{dy}{dx} = -2 - 5 \sec x. \]

### Final Answer

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = -2 - 5 \sec x. \]