Question

Consider F(x)= "integral from x to 2" (1/(1+t^4)). Find F'(x).

Answer 1

\[F(x)=\int_x^2\frac{1}{1+t^4}dt\]

Answer 2

Sorry, sorry, i meant to say from 2 to x but I just need a general idea of how to do this kind of problem so continue.

Answer 3

\[F'(x)=\frac{d}{dt}\int_x^2\frac{1}{1+t^4}dt=\frac{d}{dt}(F(2)-F(x))\]\]

Answer 4

\[\frac{d}{dt}(F(x)-F(2))\]

Answer 5

so in other words you're going to get F(2) which derivative is zero

Answer 6

now you have to worry about F(x) and use the chain rule

Answer 7

\[F(x)=F(x)-F(2)\]

\[F'(x)=f(x)\frac{d}{dx}x=1f(x)=\frac{1}{1+x^4}\]