Question

check out this question:
find \[f'(\dfrac{\pi}{2})\]
\[\large f(x)=\int_{0}^{g(x)} \frac{1}{\sqrt{1+t^3}}dt \], where \[\large g(x)=\int_{0}^{\cos x} [1+\sin (t^2)]dt\]

see if you can find -1 like I found or perhaps i made a mistake there...

Answer 1

i can't believe i posted this in feedback section lol
that hilarious :)

Answer 2

\[f'(x)=\frac{g'(x)}{\sqrt{1+(g(x))^3}} \\ g'(x)=-\sin(x) \cdot (1+\sin((\cos(x))^2) \\ g'(\frac{\pi}{2})=-1(1+\sin((0))=-1 \\ f'(\frac{\pi}{2})=\frac{-1}{\sqrt{1+(g(\frac{\pi}{2}))^3}} \\ g(\frac{\pi}{2})=\int\limits_{0}^{0}(blah) dt=0 \\ f'(\frac{\pi}{2})=-1 \]

Answer 3

yep i seem to get -1 too

Answer 4

also this in is math

Answer 5

according to my screen

Answer 6

oh great :)

Answer 7

i posted first in feedback hehe

Answer 8

oh

Answer 9

i kind of did it differently, let me analyze your way ^^

Answer 10

oh its the same actually, it thought i was something odd lol

Answer 11

thanks

Answer 12

saw *

Answer 13

lol the blah thing?

Answer 14

yeah :)

Answer 15

there is problem in max, min i couldnt solve yet
i will give it some more time and see if i can come up with something