If g(x) = int_1^(x^2) sqrt(t^3 + 1) dt, find g'(x).

Answer and explanation

Correct answer: C. 2x sqrt(x^6 + 1)

  1. Step 1: Apply the Fundamental Theorem of Calculus with the chain rule: d/dx [int_a^(u(x)) f(t) dt] = f(u(x)) * u'(x).
  2. Step 2: Here u(x) = x^2, so u'(x) = 2*x.
  3. Step 3: Evaluate integrand at u(x): sqrt((x^2)^3 + 1) = sqrt(x^6 + 1).
  4. Step 4: Multiply: g'(x) = sqrt(x^6 + 1) (2x) = 2x sqrt(x^6 + 1).
Answer
2x sqrt(x^6 + 1).
Why not the others: sqrt(x^6 + 1) omits the chain rule factor u'(x) = 2x; 2x sqrt(x^3 + 1) forgets to square x inside t^3; x^2 * sqrt(x^6 + 1) multiplies by u(x) instead of u'(x).

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