Question

If \( y = \tan^{-1} \sqrt{x^2 - 1} + \sinh^{-1} \sqrt{x^2 - 1} \), \( x > 1 \), then \( \frac{dy}{dx} = \)

• A. $ \frac{1}{x\sqrt{x^2 - 1}} $
• B. $ \frac{x+1}{x\sqrt{x^2 - 1}} $
• C. $ \frac{x+1}{x^2\sqrt{x^2 - 1}} $
• D. $ \frac{x}{\sqrt{x^2 - 1}} $

Hint:

When differentiating inverse trigonometric and hyperbolic functions involving composite expressions, use the chain rule and standard derivative formulas. Simplify step-by-step to ensure accuracy.

Answer 1

The Correct Option is B

Step 1: Differentiate $ y = \tan^{-1}\sqrt{x^2 - 1} + \sinh^{-1}\sqrt{x^2 - 1} $. 
Given: $$ y = \tan^{-1}\sqrt{x^2 - 1} + \sinh^{-1}\sqrt{x^2 - 1}. $$ We need to find $ \frac{dy}{dx} $. Use the chain rule and standard derivative formulas. 
Step 2: Differentiate $ \tan^{-1}\sqrt{x^2 - 1} $. 
Let $ u = \sqrt{x^2 - 1} $. Then: $$ \frac{d}{dx} \left( \tan^{-1}u \right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}. $$ For $ u = \sqrt{x^2 - 1} $: $$ \frac{du}{dx} = \frac{d}{dx} \left( \sqrt{x^2 - 1} \right) = \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}}. $$ Thus: $$ \frac{d}{dx} \left( \tan^{-1}\sqrt{x^2 - 1} \right) = \frac{1}{1 + (\sqrt{x^2 - 1})^2} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{1 + (x^2 - 1)} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{x\sqrt{x^2 - 1}}. $$ Step 3: Differentiate $ \sinh^{-1}\sqrt{x^2 - 1} $. 
The derivative of $ \sinh^{-1}u $ is: $$ \frac{d}{dx} \left( \sinh^{-1}u \right) = \frac{1}{\sqrt{u^2 + 1}} \cdot \frac{du}{dx}. $$ For $ u = \sqrt{x^2 - 1} $: $$ \frac{du}{dx} = \frac{x}{\sqrt{x^2 - 1}}. $$ Thus: $$ \frac{d}{dx} \left( \sinh^{-1}\sqrt{x^2 - 1} \right) = \frac{1}{\sqrt{(\sqrt{x^2 - 1})^2 + 1}} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{\sqrt{x^2 - 1 + 1}} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{x} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{\sqrt{x^2 - 1}}. $$ Step 4: Combine the derivatives. 
Now, combine the results: $$ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}\sqrt{x^2 - 1} \right) + \frac{d}{dx} \left( \sinh^{-1}\sqrt{x^2 - 1} \right). $$ Substitute the derivatives: $$ \frac{dy}{dx} = \frac{1}{x\sqrt{x^2 - 1}} + \frac{1}{\sqrt{x^2 - 1}}. $$ Factor out $ \frac{1}{\sqrt{x^2 - 1}} $: $$ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \left( \frac{1}{x} + 1 \right) = \frac{1}{\sqrt{x^2 - 1}} \cdot \frac{x + 1}{x} = \frac{x + 1}{x\sqrt{x^2 - 1}}. $$ Step 5: Final Answer. 
$$ \boxed{\frac{x+1}{x\sqrt{x^2 - 1}}} $$