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Expert-verified Found in: Page 298 ### Essential Calculus: Early Transcendentals

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 $$\sqrt {\tan x + \sqrt {\tan x} }$$.

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## 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 $$\sqrt {\tan x + \sqrt {\tan x} }$$. ### Want to see more solutions like these? 