If y = √(secx - 1)/(secx + 1) then find the value of dy/dx.

Steps to Solve

1. Identify the function to differentiate

We have the function given as $$ y = \sqrt{\frac{\sec x - 1}{\sec x + 1}} $$

2. Apply the chain rule

To differentiate, we will use the chain rule, which states that if (y = f(g(x))), then $$ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) $$ Here, let $$ u = \frac{\sec x - 1}{\sec x + 1} $$ Then (y = \sqrt{u}), and we need to find $$ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} $$

3. Differentiate (u)

Now, we will differentiate (u) using the quotient rule. The quotient rule states that $$ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $$

In our case, let (f(x) = \sec x - 1) and (g(x) = \sec x + 1):

• The derivative of (f(x)) is (f'(x) = \sec x \tan x)
• The derivative of (g(x)) is (g'(x) = \sec x \tan x)

So,

$$ \frac{du}{dx} = \frac{(\sec x \tan x)(\sec x + 1) - (\sec x - 1)(\sec x \tan x)}{(\sec x + 1)^2} $$

4. Simplify (\frac{du}{dx})

After substituting and simplifying, we notice that $$ \frac{du}{dx} = \frac{\sec x \tan x \cdot 2}{(\sec x + 1)^2} $$

5. Combine to find (\frac{dy}{dx})

Now substituting back into the equation for (\frac{dy}{dx}):

$$ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\frac{\sec x - 1}{\sec x + 1}}} \cdot \frac{2 \sec x \tan x}{(\sec x + 1)^2} $$

After simplifying the terms, we get:

$$ \frac{dy}{dx} = \frac{\sec x \tan x}{\left(\sec x + 1\right)^2 \sqrt{\frac{\sec x - 1}{\sec x + 1}}} $$