Answer :

Sure, let's differentiate the function [tex]\( f(x) = \ln \left(x^2 + 1\right) \)[/tex] step by step.

### Step 1: Define the function
The function we need to differentiate is:
[tex]\[ f(x) = \ln (x^2 + 1) \][/tex]

### Step 2: Apply the chain rule
To differentiate [tex]\( f(x) = \ln (g(x)) \)[/tex], where [tex]\( g(x) = x^2 + 1 \)[/tex], we use the chain rule. The chain rule states that:
[tex]\[ f'(x) = \frac{d}{dx}[\ln (g(x))] = \frac{1}{g(x)} \cdot g'(x) \][/tex]

### Step 3: Differentiate the inner function
Now, we need to find [tex]\( g'(x) \)[/tex] where [tex]\( g(x) = x^2 + 1 \)[/tex]:
[tex]\[ g(x) = x^2 + 1 \][/tex]
[tex]\[ g'(x) = \frac{d}{dx} [x^2 + 1] = 2x \][/tex]

### Step 4: Combine the results
Substitute [tex]\( g(x) \)[/tex] and [tex]\( g'(x) \)[/tex] back into the chain rule formula:
[tex]\[ f'(x) = \frac{1}{x^2 + 1} \cdot 2x \][/tex]

### Step 5: Simplify the expression
The derivative of the function [tex]\( f(x) \)[/tex] simplifies to:
[tex]\[ f'(x) = \frac{2x}{x^2 + 1} \][/tex]

So, the derivative of [tex]\( \ln (x^2 + 1) \)[/tex] is:
[tex]\[ \boxed{\frac{2x}{x^2 + 1}} \][/tex]