How do you use Part 1 of the fundamental theorem of calculus to find the derivative of:

g(x)=integral of (9*cos(t))/(t) from 3 to x^(1/2)?

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- Jun 12th 2007, 09:34 PMasnxbbyx113Fundamental Theorem of Calculus
How do you use Part 1 of the fundamental theorem of calculus to find the derivative of:

g(x)=integral of (9*cos(t))/(t) from 3 to x^(1/2)? - Jun 12th 2007, 09:43 PMqbkr21Re:
- Jun 12th 2007, 09:46 PMJhevon
this page may help a bit

you may also want to look up "the second fundamental theorem of calculus" on wikipedia or some other source

$\displaystyle g(x) = \int_{3}^{\sqrt {x}} \frac {9 \cos t}{t} dt$

$\displaystyle \Rightarrow g'(x) = \frac {d}{dt} \int_{3}^{\sqrt {x}} \frac {9 \cos t}{t}dt$

$\displaystyle \Rightarrow g'(x) = \frac {9 \cos \sqrt {x}}{\sqrt {x}} \cdot \left( \sqrt {x} \right)^{\prime}$ ........by the second fundamental theorem of calculus

$\displaystyle \Rightarrow g'(x) = \frac {9}{2} \frac {\cos \sqrt {x}}{ \sqrt {x}} \cdot x^{- \frac {1}{2}}$

$\displaystyle \Rightarrow g'(x) = \frac {9 \cos \sqrt {x}}{2x}$

EDIT: Ah, beaten by the great qbkr21!