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Find the derivative of ln(x^2 + 1)

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Answer
2xx2+1\boxed{\frac{2x}{x^2 + 1}}

Checking the result by integrating 2xx2+1\frac{2x}{x^2 + 1} via substitution gives ln⁡(x2+1)\ln(x^2 + 1), confirming our derivative is correct.

Problem: Find the derivative of f(x)=ln⁡(x2+1)f(x) = \ln(x^2 + 1).

Steps

  1. Apply the chain rule
    ddx[ln⁡(x2+1)]=1x2+1⋅ddx(x2+1)\frac{d}{dx}[\ln(x^2 + 1)] = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1)
  2. Differentiate the inner function
    1x2+1⋅(2x)\frac{1}{x^2 + 1} \cdot (2x)
  3. Simplify the expression
    2xx2+1\frac{2x}{x^2 + 1}

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