Math Problem Statement

Calculate the derivative of the function f(x) = √(3π) * ∫(from 1 to √x) cos(t^2) dt at the point x = π/3.

Solution

To solve this problem, we need to find the derivative of the function

f(x)=3π1xcos(t2)dtf(x) = \sqrt{3\pi} \cdot \int_{1}^{\sqrt{x}} \cos(t^2) \, dt

at the point x=π3x = \frac{\pi}{3}.

The function involves an integral with an upper limit that is a function of xx, so we’ll use the Fundamental Theorem of Calculus in conjunction with the chain rule.

Step-by-Step Solution

  1. Differentiate the integral expression: By the Fundamental Theorem of Calculus (Part 2), if we have a function F(x)=ag(x)h(t)dtF(x) = \int_{a}^{g(x)} h(t) \, dt, then F(x)=h(g(x))g(x)F'(x) = h(g(x)) \cdot g'(x).

    In our case: f(x)=3π1xcos(t2)dtf(x) = \sqrt{3\pi} \cdot \int_{1}^{\sqrt{x}} \cos(t^2) \, dt The function g(x)=xg(x) = \sqrt{x}, so g(x)=12xg'(x) = \frac{1}{2\sqrt{x}}.

  2. Apply the derivative: Using the rule for differentiating an integral with a variable limit: f(x)=3πcos((x)2)ddx(x)f'(x) = \sqrt{3\pi} \cdot \cos((\sqrt{x})^2) \cdot \frac{d}{dx}(\sqrt{x}) Simplifying (x)2=x(\sqrt{x})^2 = x, so: f(x)=3πcos(x)12xf'(x) = \sqrt{3\pi} \cdot \cos(x) \cdot \frac{1}{2\sqrt{x}}

  3. Evaluate at x=π3x = \frac{\pi}{3}: Substitute x=π3x = \frac{\pi}{3} into the expression: f(π3)=3πcos(π3)12π3f'\left(\frac{\pi}{3}\right) = \sqrt{3\pi} \cdot \cos\left(\frac{\pi}{3}\right) \cdot \frac{1}{2\sqrt{\frac{\pi}{3}}}

  4. Calculate each component:

    • cos(π3)=12\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
    • π3=π3=3π3\sqrt{\frac{\pi}{3}} = \frac{\sqrt{\pi}}{\sqrt{3}} = \frac{\sqrt{3\pi}}{3}

    Substitute these values: f(π3)=3π12323πf'\left(\frac{\pi}{3}\right) = \sqrt{3\pi} \cdot \frac{1}{2} \cdot \frac{3}{2\sqrt{3\pi}}

  5. Simplify the expression: f(π3)=33π43π=34=0.75f'\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3\pi}}{4\sqrt{3\pi}} = \frac{3}{4} = 0.75

Answer

The correct answer is 0.75\boxed{0.75}.

Would you like further details on any part of this solution?

Here are some related questions that might interest you:

  1. How does the Fundamental Theorem of Calculus apply to integrals with variable limits?
  2. What is the chain rule, and how is it used in calculus?
  3. How do we differentiate functions that involve square roots?
  4. Why does cos(π3)\cos\left(\frac{\pi}{3}\right) equal 12\frac{1}{2}?
  5. How can we simplify expressions involving square roots and constants?

Tip: When differentiating integrals with variable limits, always consider using the Fundamental Theorem of Calculus alongside the chain rule to account for any variable dependency in the limits.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation of Integrals
Chain Rule

Formulas

f'(x) = h(g(x)) * g'(x) where F(x) = ∫(from a to g(x)) h(t) dt
cos(π/3) = 1/2
Square root simplification √(π/3) = (√(3π)/3)

Theorems

Fundamental Theorem of Calculus (Part 2)

Suitable Grade Level

Undergraduate Calculus