Question

derivative of 2^u

Answer 1

\[\frac{ d }{ du }2^{u}=u2^{u-1}\]Do you need the whole proof or just that?

Answer 2

just this

Answer 3

I thought u had to have ln of something?

Answer 4

No, not on this function.

Answer 5

hmm so if it was \[2^sinepix\]

Answer 6

Do you mean?

Answer 7

y=2^(sine)(pi)(x)

Answer 8

hello?

Answer 9

\[\frac{ d }{ dx }2^{\sin(\Pi x)}=\sin(\Pi x)2\]

Answer 10

No sorry, thats not the answer I was asking it that was the derivative you want. The answer is:

Answer 11

yes that's the derivative but without parenthesis

Answer 12

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

Answer 13

Oh no if it is without the parenthesis then:

Answer 14

I got it thank u

Answer 15

Ok :)