ok this is a stupid question but... (1 Viewer)

[luigi]

how do u integrate and differentiate 2^x

that is : two to the power of x.

please reply
thanks

 

[kaseita]

Differentiate becomes:2^x ln2
Integrate becomes: 2^x/ln2 + C

It's not in the syllabus though I think.

Oh well here's the process for differentiation. same sort of thing for integration

y=2^x
log2 y=x

lny = x
ln2

lny=xln2
differentiate implicitly
end up with
dy=yln2dx
sub y=2^x

dy=2^x lnxdx

 

[DA]

A similar method is to write 2^x as e^{x ln 2}, then do it normally.

 

[underthesun]

Originally posted by kaseita
It's not in the syllabus though I think.

thank god.

 

[Minai]

but i swear ive seen a question like that in past HSC papers....

 

[McLake]

We had one like that in our school test once, but it's not in any past papers (that I can find ...)

 

[fillstar]

well the 'extension 1 understanding yr12 maths' book says it is in the syllabus but are rarely asked.

 

[Lazarus]

The derivative of y = a^x is in the syllabus, the integral is not.

 

[Dumbarse]

without differentiating implicitly
cause that not 3u.

y = 2^x

log y = log 2^x

log y = x log2

y = e^xlog2

dy/dx = log2 e^xlog2
??

i dunno , how would u do it??

 

[McLake]

Originally posted by Dumbarse
without differentiating implicitly
cause that not 3u.

y = 2^x

log y = log 2^x

log y = x log2

y = e^xlog2

dy/dx = log2 e^xlog2
??

i dunno , how would u do it??

Here you go Dumbarse:

y = a^x
y = [e^(lna)]^x
y = e^(xlna)
dy/dx = lna * e^(xlna)
dy/dx = lna * a^x

Ask for the intergration if you want ...

 

[macca202]

sex and candy i think.....

 

[macca202]

hey dont u hate it when you spend like ages trying to get into the forum to see what the last reply was, and to only find out that the person just said crap, and so because of this one insignificant person, you wasted time that should have been spent doing 3unit past papers. :mad1:

 

[Lazarus]

Originally posted by Dumbarse
dy/dx = log2 e^xlog2
??

i dunno , how would u do it??

That's correct, assuming you're using base e logs... and:

log(2).e^x.log(2)
= log(2).e^{log(2^x)}
= log(2).2^x