Math Problem Statement
Solution
To solve this problem, we need to find the derivative of the function
at the point .
The function involves an integral with an upper limit that is a function of , so we’ll use the Fundamental Theorem of Calculus in conjunction with the chain rule.
Step-by-Step Solution
-
Differentiate the integral expression: By the Fundamental Theorem of Calculus (Part 2), if we have a function , then .
In our case: The function , so .
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Apply the derivative: Using the rule for differentiating an integral with a variable limit: Simplifying , so:
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Evaluate at : Substitute into the expression:
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Calculate each component:
Substitute these values:
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Simplify the expression:
Answer
The correct answer is .
Would you like further details on any part of this solution?
Here are some related questions that might interest you:
- How does the Fundamental Theorem of Calculus apply to integrals with variable limits?
- What is the chain rule, and how is it used in calculus?
- How do we differentiate functions that involve square roots?
- Why does equal ?
- 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
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