Question

i hav a questien about calculs

Answer 1

no pls

Answer 2

I can help

Answer 3

\[\frac{d}{dx}(\log_xx^x)\]

Answer 4

oh gosh um wow sorry i thought i could but i dont even know where to start with that.

Answer 5

1

Answer 6

ya rite no way

Answer 7

is 1 the answer

Answer 8

no

Answer 9

wate batmans rong

Answer 10

its differenciated
and explain ur answers
sigh

Answer 11

it is 1

Answer 12

hahaha Batman i could see him saying that "NO" lol

Answer 13

\[\log _{x}(x^x) = \frac{ logx^x }{logx}\]

Answer 14

is it 1 or not

Answer 15

NO IT'S NOT 1

Answer 16

Guys your making him mad

Answer 17

Use quotient rule now

Answer 18

no batman ur rong

Answer 19

\[\frac{logx^x}{logx}=\frac{xlogx}{logx}=x\]

Answer 20

!!!!

Answer 21

\[\huge \frac{ d }{ dx }\frac{ u }{ v }=\frac{ v \frac{ du }{ dx }-u \frac{ dv }{ dx } }{ v^2 }\]

u = log(x^x), v = log(x)

Answer 22

Go batman go!
*confetti*

Answer 23

\[d/dx(x)=1\]

Answer 24

lol

Answer 25

log rules batmen drules

Answer 26

her english is giving me heart attack
but she is quite correct abt it
basic logarithm rule x is the base,and its raised to the powe x to give x^x so log x^x to base x is x,so x differentiated gives 1

Answer 27

sry my natiff langich is russien but i no sum germen

Answer 28

^dat hawt

xD

Answer 29

lol

Answer 30

yusssssss! im sexey on the interweb

Answer 31

bimbo

Answer 32

\[\implies \frac{ -\log(x^x)(\frac{ d }{ dx }(logx))+\log(x)(\frac{ d }{ dx } (\log(x^x))}{ \log^2x }\]

\[ = \frac{ \log(x)(\frac{ d }{ dx }(\log(x^x)) -\frac{ 1 }{ x }(\log(x^x))}{ \log^2)(x) }\]

\[= \frac{ \frac{ -\log(x^x) }{ x }+\log(x)\frac{ d }{ dx }(x)+x \frac{ d }{ dx }logx(\log(x)) }{ \log^2(x) }\]

\[= \frac{ \frac{ -\log(x^x) }{ x }+\log(x)(\log(x)+\frac{ 1 }{ x }x }{ \log^2(x) } \implies ~ \frac{ \log(x)(1+\log(x))-\frac{ \log(x^x) }{ x } }{ \log^2(x) }\]

Answer 33

batman -.-

Answer 34

seriously?

Answer 35

Cry

Answer 36

How about I give you love? <3

Answer 37

lool

Answer 38

There's your answer, now it's time to go save Gotham.

Answer 39

that's what i have been saying since time immemorial it is 1

Answer 40

**highfives nobody**

Answer 41

You guys are killing me

Answer 42

poor batmen using qotient rule. i jest use prodccut rule n chane rool evry time. \[f/g = fg^{-1}\]

Answer 43

http://24.media.tumblr.com/tumblr_mah6v1A5Xp1rziwwco1_250.gif me right now, you're all bullies.

Answer 44

**slow clap**