Question

Take the derivative of log(base 2)x=(2^y)+(x^2)(y^2)

Answer 1

no, dont take the derivative. be a non conformist. take the integral for gods sake!

Answer 2

problem says take derivative

Answer 3

haha ok you have two methods of doing this, one is easier than the other. the first is to just take the derivative, in which case you need to memorize a formula, and the other takes hardcore algebra. ill cover the first

Answer 4

i have no idea O.o

Answer 5

is the question to find dy/dx ???

Answer 6

yes

Answer 7

needs both implicit and dealing with logs

Answer 8

ok... do you know implicit differentiation?

Answer 9

yes

Answer 10

\(\huge log_2x=2^y+x^2y^2 \)
\(\huge [log_2x]'=[2^y]'+[x^2y^2]' \) take the derivative of both sides...
can you find all those derivatives?

Answer 11

how do you take derivative of \[\log_{2}x \]

Answer 12

1/xln2

Answer 13

yes ^^^. you can get that by formula or just obtain it by change ofbase...

Answer 14

how do you do the 2^y tho?

Answer 15

i feel like i should know this -.- :/

Answer 16

ok so for the 2 to the y do i also do ln

Answer 17

\(\large [2^y]'=ln2 \cdot 2^y \cdot y' \)
hehe... formula...

Answer 18

... along with the chain rule of course...

Answer 19

i feel embarrassed

Answer 20

thats exactly what i had gotten, then ditched because i thought it was wrong

Answer 21

that last one is using the product rule with the chain rule: \(\large [x^2y^2]'= \) ???

Answer 22

so setting 2^y=w and then solving for dw= (you get 2^y * ln2 *y') is sufficiently how you would prove that formula for d 2^y ?

Answer 23

yeah.. ^^^

Answer 24

of course

Answer 25

so after u got all that... do some algebra isolate the y'....

Answer 26

ok

Answer 27

yeah its simple implicit after that. id say the hardest part was recognizing that you can change the base in the 2^y.

Answer 28

sorry... i have to make like a baby.... and head out.....
cya... :)

Answer 29

hwallday, you should have gotten something like ((1/xln2)-2xy^2)/((2^y)ln2+2yx^2)

Answer 30

yup