This has to be 20 Characters long to tell you the answer, but the answer is A
Hello from MrBillDoesMath!
Answer:
(3 /2)* ( x^(1/2) )
Discussion:
I think you are asking for the derivative of (3x) ^ (1/2) -- the square root of both 3 and x.
The derivative of (3x) ^ (1/2) is
(1/2) (3x) ^ (1/2 -1) (d(3x)/dx) =
(1/2) (3x)^(-1/2) 3 =
3/2 (1/ (3x)^ (1/2) ) =
3/2 ( 1/ 3^(1/2)) ( 1 / x^(1/2) ) =
(3/3^(1/2)) * (1/2) ( 1/ x^(1/2) ) =
( 3^(1/2) / 2 ) * (1/ x^(1/2) ) =
( 3^(1/2) ) * (1 / 2) * (1/ x^(1/2) ) =
It's straightforward but using a keyboard over complicates its appearance
Thank you,
MrB
Given
Collecting the like terms
The coefficient of
is 10
The constant is 3
Answer:
x=square root 63. Basically the radical 63.
Step-by-step explanation:
So in the first triangle with the 3 in it. You have root 3 and use the a^2 +b^2=c^2 formula. So 3^2 + 3^2= x^2 and then when you solve you get radical 18^2 which is equal to radical 18 or square root 18. Then we go to the bigger triangle and we do the same formula but this time we have the c value which is 9. So we do radical 18^2 + x^2 = 9^2 and then solve that and minus 18 from the 81 since the radical goes off the 18 and you get square root63.
Sorry i couldn't use the signs. Sorry If I'm wrong Hope this helps!.
use sin^2(x) =1/2 * (1-cos2x)
cos 2x - use def cosx from table
f(x)=sin^2(0)
f'(x)=2*sin(x)*cos(x)
therefore f'(x)=sin(2x)
f''(x)=2cos(2x)
f'''(x)=-4sin(2x)=-4*f'(x)
<span>putting it into maclaurin's series</span>
<span>We get,</span>
<span> 0+0+ 2x^2/2! + 0...</span>