How do I find f'(x)? When f(x)=∫dt/√(1+t^2)?
f(x)=∫dt/√(1+t^2)
(integral from 0 to g(x))
g(x)=∫(sint-1)
(integral from 0 to cosx
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f(x)=∫dt/√(1+t^2)
(integral from 0 to g(x))
g(x)=∫(sint-1)
(integral from 0 to cosx
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INTEGRAL CALCULUS
For the fundamental calculus' theorem :
... h(x)
D { ∫ [ f ( t )·dt ] } = h'(x)·f ( h(x) ) - g'(x)·f ( g(x) )
... g(x)
Example :
F ( x ) = ∫[ 0 , cos( x ) ] { [ sin( t ) - 1 ]·dt } ;
F ' ( x ) = D[ cos( x ) ]·[ sin( cos( x ) ) - 1 ] - D( 0 )·[ sin( 0 ) - 1 ]
.......... = sin( x )·[ 1 - sin( cos( x ) ) ]
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