Question

find dy/dx for y=cos(ln2x^2) CLICK to see my answer, but I do not know if it is right...

Answer 1

Product rule: cos(1/2x^2) + (ln2x^2)(-sin)

Answer 2

is this the original...

Answer 3

No, that's as far as I have gotten. but i do not know if i am on the right track...

Answer 4

\[y=\cos(\ln2x ^{2})\]

Answer 5

no i mean that ^ is that the original?

Answer 6

^yes that is the original problem

Answer 7

then no your wrong

Answer 8

ln2x^2 is the angle

Answer 9

it's not being multiplied with cosine.

Answer 10

for you to do product rule it would have to look like this....

Answer 11

\[\cos(x)\ln(2x ^{2})\]

Answer 12

oh! so is this the chain rule?

Answer 13

yes

Answer 14

start with the outside leave the inside alone, then take derivative of inside.

Answer 15

so: -sin(ln2x^2)(1/2x^2) ?

Answer 16

almost

Answer 17

that last part is 1/(2x^2)

Answer 18

missing something

Answer 19

what is the derivative of lnx?

Answer 20

1/x

Answer 21

so 1/(2x^2) ?

Answer 22

your really close here let me give you a hint and i think your gonna get it.