Question

Find an equation for the derivative of the given function: y=2^x lnx
The answer should be -sin(lnx)(1/x). Can someone show me how to get there? Thanks!!

Answer 1

use the product rule
f'(x) = 2^x * 1/x + lnx * f'(2*x)

let y = 2^x
lny = x ln2
1/y dy/dx = ln 2
dy.dx = 2^x * ln2

f'(x) = 2^x * 1/x + lnx* 2*x* ln x
= 2^x (1/x + (lnx )^2)

- not quite what you have ! i might have missed something here

Answer 2

I'm sorry, I wrote down the answer to the wrong problem. The answer should be 2^xlnx[ln2+1/(xlnx)]

Answer 3

Or if this is more clear: (2^x)lnx[ln2+1/(xlnx)]

Answer 4

oh i made an error in the last step
its f'(x) = 2^x * 1/x + ln2 * 2^x * ln x
= 2^x 2^x ln2 lnx
--- +
x

= 2^x + 2^x ln2 x lnx
-----------------
x
= 2^x ( 1 + ln2 x lnx )
-----------------
x

Answer 5

- cant seem to get your answer - i'd better try wolfram alpha

Answer 6

Thank you so much for your help! I'll try to figure the rest out myself! Thank you!!!

Answer 7

wolfram gives same answer as me
your welcome