Question

If $ y = \ln(x^2 + 1) $, then find $ \frac{dy{dx} $ at $ x = 1 $.

• A. $ \frac{1}{2} $
• B. $ \frac{1}{3} $
• C. $ 1 $
• D. $ \frac{2}{3} $

Hint:

When differentiating a logarithmic function, use the chain rule: $ \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} $.

Answer 1

The Correct Option is C

Step 1: Differentiate $ y = \ln(x^2 + 1) $.
The function is: $$ y = \ln(x^2 + 1). $$ Using the chain rule, the derivative is: $$ \frac{dy}{dx} = \frac{d}{dx} \left[ \ln(x^2 + 1) \right] = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1). $$ Compute $ \frac{d}{dx}(x^2 + 1) $: $$ \frac{d}{dx}(x^2 + 1) = 2x. $$ Substitute this back: $$ \frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}. $$ Step 2: Evaluate $ \frac{dy{dx} $ at $ x = 1 $.}
Substitute $ x = 1 $ into the derivative: $$ \frac{dy}{dx} \bigg|_{x=1} = \frac{2(1)}{1^2 + 1} = \frac{2}{1 + 1} = \frac{2}{2} = 1. $$ Conclusion: The value of $ \frac{dy}{dx} $ at $ x = 1 $ is: $$ \boxed{\text{C) } 1} $$