Question

how would you find d^2y/dx^2 of:
y=sin^2(2x)?

Answer 1

\[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ d }{ dx }(\frac{ dy }{ dx })\] Do you know how to derivate?

Answer 2

yep, i first tried to find the 1st derivative of sin^2 2x and got 4sin2xcos2x but can't seem to find the derivative of that :/

Answer 3

You can find the derivative of 4sin2xcos2x by using this rule : d/dx(f*g)=f*d/dx(g)+g*d/dx(f)

Answer 4

yes math is right
this is called the product rule used to differentiate the product of 2 functions

so you need to differentiate 4sin2x and cos2x - can you do that?

Answer 5

i thought so but i keep getting the wrong answer :(
would you do 4sin2x x -2sin2x + cos2x x 8cos2x?

Answer 6

thats correct

Answer 7

it can be simplified

Answer 8

thanks, but do you know how you would get this to 8cos4x? which is the answer I'm meant to end up with :)

Answer 9

ok - i think you use the identity

cos2x = cos^2x - sin^2x

if we replace x by 2x we get

cos4x = cos^2 2x - sin^2 2x

Answer 10

CALCULATOR BRAH 2ND DEGREE SOMETHING LIKE THAT AND PUT SIN AND TYPE THAT IN BOOOOM

Answer 11

and we can simplify your result to 8( cos^2 2x - sin^2 2x)

do you see the connection?

Answer 12

8( cos^2 2x - sin^2 2x) = 8 cos 4x

Answer 13

yes, thank you very much! :D