Question

Take the derivative of y=x^(2x)

Answer 1

logarithmic differentiation

Answer 2

Is the answer 2x(x)^(2x-1) *(2)

Answer 3

ah, no then

Answer 4

yep,
since you have exponent as variable, you'll have to take log on both sides first

Answer 5

y = x^(2x)
ln y = ...?

Answer 6

hmm

Answer 7

I forgot all these rules :o I think the bases switch

Answer 8

maybe 2x(x) ?

Answer 9

I need to look over these logs >_<

Answer 10

\(\Large \ln a^b = b \ln a\)

Answer 11

2xlnx then?

Answer 12

yesss

ln y = 2x ln x
now differentiate both sides w.r.t x

Answer 13

ok

Answer 14

chain rule on left side
product rule on right side

Answer 15

ok so 1/x on the left?

Answer 16

there's no 'x' on left :P

Answer 17

hmm wait

Answer 18

oh I'm hopeless with logs

Answer 19

I'm stuck

Answer 20

\(\Large d/dx (\ln y) = (1/y) d/dx (y) = (1/y)(dy/dx)\)

Answer 21

So this rule applies, it's just not x, it's y

Answer 22

got that ?
and what about right side ?

Answer 23

doing it

Answer 24

(2)(lnx)+(2x)(1/x)

Answer 25

which simplifies to 2lnx+2 if it's correct??

Answer 26

correct!

Answer 27

now just isolate dy/dx

Answer 28

by multiplying y on both sides

Answer 29

and then plugging in y = x^(2x)

Answer 30

just so it's clear: currently (1/y)(dy/dx)=2lnx+2 ?

Answer 31

correct

Answer 32

ok then

Answer 33

to isolate dy/dx ,what u need to do ??

Answer 34

multiply by y

Answer 35

dy/dx=y(2lnx+2)

Answer 36

y=x^2x

Answer 37

yes., go on, and factor out the 2 too

Answer 38

so dy/dx=(x^2x)(2lnx+2)=

Answer 39

factor the 2, and you'll be done :)

Answer 40

(x^2x)(2)(lnx+1)

Answer 41

thats it! correct :)

Answer 42

whew, thanks :)

Answer 43

welcome ^_^

Answer 44

are logs very common in upper level calculus?

Answer 45

not very common, but you will see them here and there.....always good to know them properly.

Answer 46

ok, thanks for the advice