Question

dy/dx
answer check!

Answer 1

y=log(logx)+2^sinx

Answer 2

\[\log(logx)=> \frac{1}{logx} \times \frac{1}{x} = \frac{1}{xlogx} \]

Answer 3

\[\LARGE u=2^{sinx}\]..retriception

Answer 4

wtf assumption*

Answer 5

\[\frac{1}{u} \times \frac{du}{dx} = \log2cosx\]

Answer 6

1/(x log x) is correct.

Answer 7

\[\frac{du}{dx}=2^{sinx} \times \log2cosx\]

Answer 8

whats the derivative of a^x ?

Answer 9

then combine o.o

Answer 10

where is a?

Answer 11

if a is a constant then should be 0 otherwise xloga

Answer 12

a^x log a

Answer 13

\[\LARGE logu=sinxlog2\]

Answer 14

so, for 2^sin x
it will be
2^sin x log 2 d/dx(sin x)
chain rule.
no need of log. diff.

Answer 15

why cant we do log diff

Answer 16

with log diff, u get the same answer actually.

Answer 17

exactly

Answer 18

WFA certified

Answer 19

i never said, you cannot do, i said its not required.

Answer 20

so its not incorrect either right?

Answer 21

its not incorrect...just a longer way

Answer 22

i didnt understand what u did without log one^

Answer 23

then do log :P

Answer 24

yeah better o.o thanks :D

Answer 25

and get the same answer

Answer 26

okay :o
IM STARTING TO LOVE THAT METHOD SINCE YESTERDAY <3

Answer 27

lol....nice..thats what practice odes to any1..

Answer 28

*does

Answer 29

:D