Question

If f is the function f(x)=Integral b=2x a=4
(t^2-1)^(1/2) then f'(2)=

Answer 1

Is the question:
\[\int\limits (t^2-1)^{\frac{ 1 }{ 2 }} dt\]

Answer 2

oops I mean \[(t^2-t)^(1/2)\] with the integral from 2x to 4. 2x at top. 4 at bottom

Answer 3

\[f(x)=\int\limits_{4}^{2x}(t^2-1)^{1/2}dt\]

So since you're really just integrating and taking the derivative again, what you can do is just replace (t) with (2x).

Answer 4

square root ot t^2-T

Answer 5

\[\int\limits_{4}^{2x}(t^2-t)^{\frac{ 1 }{ 2 }} dt\]

You basically use FToC (fundamental theorem of calculus). Plug in 2x for t and use chain rule.

Answer 6

The answer isn't a variable
a)0 b) 7/2sqr12 c sqr2 d)sqr12 e)2sqr12

Answer 7

Did you tell us everything?

Answer 8

Yep that's the problem.

Answer 9

So it looks exactly like one of these (look at the dx or dt)

\[f(x)=\int\limits_{4}^{2x}(t^2-t)^{1/2}dt\] or \[f(x)=\int\limits_{4}^{2x}(t^2-t)^{1/2}dx\]

Answer 10

Yes trying to figure out the value of f'(2) using this info.

Answer 11

Don't I cancel the integral with the derv and just plug in the 2?

Answer 12

I assumed you use fundamental theorem of calculus to get the derivative.
then plug in 2.

Answer 13

I think the integral and dervivative cancel out because I'm trying to find the derv.

Answer 14

I got e) 2sqrt(12) because you must also remember that upon taking the derivative you have to use the chain rule.

See, when you integrate you have f(x). So when you take the derivative you must multiply it by the derivative of the inside, which is 2.

\[f(x)=\int\limits_{4}^{2x}(t^2-t)^{1/2}dt\]\[f'(x)=((2x)^2-2x)^{1/2}*2\] Now plug in 2.

Answer 15

Me too. So is the integral from 2x to 4 not used in the problem?

Answer 16

No.