Question

dy/dx =

Answer 1

\[y=x^{\log_4(x)}\]

Answer 2

Take the derivative with respect to x.

Answer 3

lny = lnx^log_4x

Answer 4

\(ln(y)=log_4(x)ln(x)\)

Answer 5

lny = (log_4x)(lnx)

Answer 6

correct

Answer 7

then how do i derive logs with base log 4?

Answer 8

is it still 1/x?

Answer 9

\((\log_a(x))'=\frac{1}{xln(a)}\)

Answer 10

wait is it logx/log4 = logx - log4?

Answer 11

no

Answer 12

log(a/b) = ...

Answer 13

ahh yeah

Answer 14

\(\frac{1}{xln(4)}\)

Answer 15

so 1/xlog4

Answer 16

correct

Answer 17

i see then use the product rule and multiply by y

Answer 18

yep

Answer 19

thanks ;)

Answer 20

chain rule on the left...

Answer 21

?

Answer 22

I get

\(x^{\log_4(x)}[\frac{ln(x)+ln(4)\log_4(x)}{xln(4)}]\)

Answer 23

\(ln(y)=\log_4(x)ln(x)\) then we take the derivative of both sides

\(\frac{1}{y}\frac{dy}{dx}=\frac{1}{xln(4)}ln(x)+\frac{1}{x}\log_4(x)\\\) so
\(\frac{dy}{dx}=y(\frac{ln(x)}{xln(4)}+\frac{\log_4(x)}{x})\) now plug in for y
\(\frac{dy}{dx}=x^{\log_4(x)}(\frac{ln(x)}{xln(4)}+\frac{\log_4(x)}{x})\)

Answer 24

I think

Answer 25

hmmm

Answer 26

oh i see it

Answer 27

http://www.wolframalpha.com/input/?i=%282x^%28log%28x%29%2Flog%284%29-1%29log%28x%29%29%2Flog%284%29%3D+x^ {\log_4%28x%29}%28\frac{ln%28x%29}{xln%284%29}%2B\frac{\log_4%28x%29}{x}%29

Answer 28

the one on the left is the answer the wolfram gives for the derivative, and the one on the right is the answer I gave

Answer 29

holy toledo, this is intense

Answer 30

got to love wolframalpha:)

Answer 31

after that i had to convert to natural logs, yeah haha my homework is online so i have to get everything PERFECT for it to accept the answer, lol

Answer 32

but i got it :p

Answer 33

:)

Answer 34

good question:)

Answer 35

thanks haha

Answer 36

you do good in math?

Answer 37

yeah I'm taking my 2nd of 4 calculus courses

Answer 38

I had a feeling....you seem like you got it.

Answer 39

gonna be a genius when I'm done!

Answer 40

think about a math minor, it will look good....

Answer 41

im gonna major, or at least try

Answer 42

oh sweet

Answer 43

nm then... you will be fine

Answer 44

:D