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Q11E

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Essential Calculus: Early Transcendentals
Found in: Page 298
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find the derivative of the function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\) using Part 1 of The Fundamental Theorem of Calculus.

The derivative of function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\) is \(\sqrt {\tan x + \sqrt {\tan x} } \).

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Step by Step Solution

Step 1: The Fundamental Theorem of Calculus, Part 1.

If \(f\)is continuous on \({\rm{(a, b)}}\), then the function \({\rm{g}}\) is defined by \(g(x) = \int_a^x f (t)dt,a \le x \le b\)is an anti-derivative of \(f\), that is, \({g^\prime }(x) = f(x)\)for \({\rm{a < x < b}}\).

 Step 2: Use The Fundamental Theorem of Calculus for calculation.

The given function is \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\).

Obtain the derivative of the function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\), using the above-mentioned theorem as follows.

As per the fundamental theorem of calculus, if for \(a\) continuous function \(f\), \(g(x) = \int_a^x f (t)dt\), then \({g^\prime }(x) = f(x)\).

Therefore, by comparing the given function with the theorem, it is obtained that, \({\rm{g(x) = y}}\) and \(f(t) = \sqrt {t + \sqrt t } \). Then, \({y^\prime }(x) = f(x)\).

Therefore, by applying the above-mentioned theorem, the derivative of \(y\)is obtained as \({y^\prime }(x) = \sqrt {\tan x + \sqrt {\tan x} } \).

Therefore, the derivative of function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\) is \(\sqrt {\tan x + \sqrt {\tan x} } \).

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