Question

find f'x for

Answer 1

\[f(x) = \int\limits_{2}^{x^2}(1/t ^{2})dt\]

Answer 2

1/x^2. This is the 2nd rule of the Fundamental Theorem of Calculus I believe.

Answer 3

thats what im guessing as well, but is there anything else to it, with all the intervals?

Answer 4

or are they just here to confuse me.

Answer 5

Nope, nothing else needed. That's all that it is to find it. They are there to confuse you :).

Answer 6

Make it easier to read\[\Large \frac{ d }{ dx } \int\limits\limits_{2}^{x^2}\frac{ 1 }{ t^2 }dt\]

Answer 7

@derpfacelel it's not 1/x^2
\[\Large \frac{ d }{ dx } \int\limits\limits\limits_{2}^{x^2}\frac{ 1 }{ t^2 }dt = \frac{ 1 }{ (x^2)^2 }\]

Answer 8

wouldnt it be 1/x^2, because f(x) is equal to the integral of f'(x) so the answer would be right there

Answer 9

it is 1/x^4, was mistaken. wasn't careful

Answer 10

Oh, wait, I forgot you need to multiply by the derivative of x^2 \[\Large \frac{ d }{ dx } \int\limits\limits\limits\limits_{2}^{x^2}\frac{ 1 }{ t^2 }dt = \frac{ 1 }{ (x^2)^2 }*2x\]

Answer 11

http://ltcconline.net/greenl/courses/105/antiderivatives/secfund.htm

Scroll down to this image: http://ltcconline.net/greenl/courses/105/antiderivatives/secfun4.gif

Answer 12

@zubhanwc3 @derpfacelel I made a mistake too.