d/dx of the integral ??

[eamon7]

how to calculate :

\(\displaystyle \dfrac{d}{dx}\, \int_0^1\, \dfrac{2^t}{ \log(x + 2^t )}\,dt\)

the bound for the integ its from [0,1]

thx

 

[Steven G]

how to calculate :


[MATH] d/dx ∫ 2^t/ log(x + 2^t )dt [/MATH]
the bound for the integ its from [0,1]


thx

Let u = x + 2^t. Then see what to do next.

EDIT: I did not notice the d/dx in front

 

[Ishuda]

how to calculate :


[MATH] d/dx ∫ 2^t/ log(x + 2^t )dt [/MATH]
the bound for the integ its from [0,1]


thx

The definite integral of a function f(x,t) with respect to t is going to be some function of x, that is
\(\displaystyle \int_a^b f(t,x) dt = F(t,g(x))|_a^b\)
where F(t,g(x)) is the indefinite integral. Now we can just use the chain rule.

As a simple example [I don't know what the above integral is but it really doesn't matter], we have
\(\displaystyle \int_a^b cos(x^2 t) dt = sin(b x^2) - sin(a x^2)\)
and we can write g(x)=x² with
F(t,g(x))\(\displaystyle |_a^b\) = sin(t g(x))\(\displaystyle |_a^b\)
So, if we want the derivative of F, we can use the chain rule
d(F\(\displaystyle |_a^b\))/dx = (dF/dg)\(\displaystyle |_a^b\) dg/dx
Well dF/dg is just f(x,t) and we have
\(\displaystyle \frac{d(\int_a^b cos(x^2 t) dt)}{dx} = 2x ( cos(b x^2) - cos(a x^2) )\)