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Q11E

Expert-verifiedFound in: Page 298

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**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 \(\sqrt {\tan x + \sqrt {\tan x} } \).

**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 **\({\rm{a < x < b}}\)**.**

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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