`y = int_1^(sin (x)) (sqrt(1 + t^2))dt` Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

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Chapter 5, 5.3 - Problem 18 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the Part 1 of the FTC to evaluate the derivative of the function. You need to notice that the function h(x) = y is the composite of two functions `f(x) = int_1^x sqrt(1+t^2)dt` and g(x) = sin x, hence `h(x) = f(g(x)).`

Since, by FTC, part 1, `f'(x) = sqrt(1+x^2)` , thenĀ  `h'(x) = f'(g(x))*g'(x).`

`h'(x) = sqrt(1+sin^2 x)*cos x`

Hence, evaluating the derivative of the function, using the FTC, part 1, yields `y = sqrt(1+sin^2 x)*cos x.`

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