If \(h(x) = \ln(x^2 + 1)\), find \(h'(x)\).

["### Differentiating ( h(x) = \ln(x^2 + 1) ): A Complete Guide", "Understanding how to differentiate composite functions is a fundamental skill in calculus. If you're working with the function ( h(x) = \ln(x^2 + 1) ), finding its derivative ( h'(x) ) is essential for applications in optimization, curve analysis, and beyond. This article explains step-by-step how to compute the derivative of ( h(x) ), shedding light on key concepts like chain rule and logarithmic differentiation.", "---", "### What is ( h(x) = \ln(x^2 + 1) )?", "The function ( h(x) = \ln(x^2 + 1) ) combines a logarithmic function with a quadratic expression inside the logarithm. Here:", "- The inner function is ( u(x) = x^2 + 1 ).\n- The outer natural logarithm part is ( \ln(u) ).", "This composition means ( h(x) ) is a composite function, and we apply the chain rule to differentiate it efficiently.", "---", "### Applying the Chain Rule to Find ( h'(x) )", "The Chain Rule states:", "If ( h(x) = \ln(u(x)) ), then\n[\nh'(x) = \frac{d}{dx}[\ln(u(x))] = \frac{1}{u(x)} \cdot u'(x).\n]", "#### Step 1: Identify ( u(x) )\nLet\n[\nu(x) = x^2 + 1.\n]", "Then the derivative of ( u(x) ) is:\n[\nu'(x) = 2x.\n]", "#### Step 2: Apply the Chain Rule\nSubstitute into the chain rule formula:\n[\nh'(x) = \frac{1}{u(x)} \cdot u'(x) = \frac{1}{x^2 + 1} \cdot 2x.\n]", "Thus, the derivative is:\n[\nh'(x) = \frac{2x}{x^2 + 1}.\n]", "---", "### Why This Matters: Applications of ( h'(x) )", "Computing ( h'(x) = \frac{2x}{x^2 + 1} ) is not just an academic exercise. This derivative allows you to analyze how the original function ( h(x) = \ln(x^2 + 1) ) behaves:", "- Critical points: Setting ( h'(x) = 0 ) helps locate maxima, minima, or points of inflection in curves representing ( h(x) ).\n- Graph behavior: The sign and magnitude of ( h'(x) ) describe where ( h(x) ) is increasing or decreasing.\n- Optimization: Use derivatives to maximize or minimize quantities modeled by ( h(x) )—common in economics, physics, and engineering.", "---", "### Final Answer", "[\n\boxed{h'(x) = \frac{2x}{x^2 + 1}}\n]", "---", "### Quick Recap", "| Step | Explanation |\n|-------------------------------|-------------------------------------------------|\n| Identify inner function (u(x)) | (u = x^2 + 1) |\n| Differentiate outer function | Derivative of (\ln(u)) is (1/u \ imes u') |\n| Compute (u') | (u' = 2x) |\n| Combine using chain rule | (h'(x) = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}) |", "---", "By mastering differentiation like this, you unlock deeper insights into functions involving logarithms and polynomials—key tools in calculus. Practice computing derivatives of similar composite functions to build confidence and precision.", "Keywords: compute ( h'(x) ), derivative of ( \ln(x^2 + 1) ), chain rule, calculus tutorial, function differentiation, ( h(x) = \ln(x^2 + 1) ), mathematical steps."]