Calculus. Please help me and I will give you medals.
@loser can you help me?
@mathmale @freckles
i used the calculator and got it, is that all you need to know is how to type it in
but that's at the second derivative. They gave me the original function. Then I need to derive it to cancel the integral and then I need to drive it again to find the second derivative, but I need help on the limits of this function. How do would you plug the limits after you derive the integral?
derive*
I got sec(4x^2)*8x as the second derivative, but it doesn't give me any of the values.
I know you can just plug it in the calculator, but I need to know how to get the answer
what did you get for F'(x)?
hint use fundamental theorem of calculus
F ' (x) = 2*tan(4x^2)
I got, tan(4x^2)
how did you get the 2?
and you didn't find the derivative of the limit there... \[F(x)=\int\limits_0^{a(x)}g(t) dt \\ F(x)=G(t)|_0^{a(x)} \\ \text{ assume } G'=g \\ F(x)=G(a(x))-G(0) \\ \text{ differentiate both sides }\]
I used chain rule for the first term and constant rule for the second term
\[ F'(x)=[G(a(x))]'-[G(0)]' \\ F'(x)=a'(x)G'(a(x))-0 \\ F'(x)=a'(x)g(a(x))\]
okay I see. I forgot the chain rule.
oh isee, calc for the decimal approx answer, i figured just type the thing into a computer algebra system and hit enter.. hah
i know it, but I forgot to apply it
so @DanJS has actually given you the correct F' and now F'' should be easy to find
i got sec^2(4x^2)*16x
oh and actually your earlier F'' @Joseluess was almost there you just needed a multiple of 2 there so great job and this F'' is like totally right now
you will definitely get one of the answers after plugging in 1
make sure your calc is in radians
16.078 and yes it is. Thank you
no no
your calc is in degrees
put in radians you can do this by choosing mode
wait, it is in degrees
37.44
cool stuff
yeah radians is like 37.4
is that correct?
yep you are right I got like 37.4488blahblah
haha thanks both of you goys.
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