differentiation (1 Viewer)

[micuzzo]

^meant to say integration



hey can someone plz explain how i would ANTIdifferentiate...ok i mean integrate 5/3x

i dont understand how to use f'(x)/f(x) dx = ln f(x) + C

 

[lyounamu]

hey can someone plz explain how i would differentiate 5/3x

i dont understand how to use f'(x)/f(x) dx = ln f(x) + C

uh...

y = 5/3x = 5/3 x^-1

dy/dx = -5/3 x^-2 = -5/3x^2

But I think you mean integration by looking at your second question....

y' = 5/3x = 5/3 . 1/x
y = 5/3 lnx +c

 

[lyounamu]

derivative of ln x = 1/x + c

derivative of ln f(x) = f'(x)/f(x) + c

Apply to question and go.

So, to differentiate 5/3x, you need a 3 on top (3 = derivative of 3x).

Thus, you manipulate to look like this:

5/3 x (3/3x)

The use the rules above to end up with:

[5 ln (3x)]/3

You just did integration...

 

[lychnobity]

hence, the deleted post

 

[micuzzo]

uh...

y = 5/3x = 5/3 x^-1

dy/dx = -5/3 x^-2 = -5/3x^2

But I think you mean integration by looking at your second question....

y' = 5/3x = 5/3 . 1/x
y = 5/3 lnx +c

yar im a lost dude... i meant integration...will fix

but i still dont get it

 

[micuzzo]

cant it b 3/5 ln 3x +C

 

[lolokay]

well it could be 5/3 ln 3x +C (not 3/5)
this is equivalent to 5/3 ln x +C, since ln 3x = ln x + ln3, and ln3 is just a constant

 

[lyounamu]

lol, sorry for not explaining straight.

 

[micuzzo]

so let me get this straight... we need to make the numerator the same as the derivative of the denominator right..?

so cant we just say 3/5 S 5/3x = 3/5 ln 3x + C [where S is the integrand sign] (since 3/5 x 5 = 3)

please correct me someone...

 

[lyounamu]

so let me get this straight... we need to make the numerator the same as the derivative of the denominator right..?

so cant we just say 3/5 S 5/3x = 3/5 ln 3x + C [where S is the integrand sign] (since 3/5 x 5 = 3)

please correct me someone...

should be 5/3 ln3x + c

think about it, you go from 5/3x to 5/3 . 3/3x

 

[micuzzo]

yar ok still lost...

 

[lolokay]

at the second line you could also use 1/x, or anything equivalent

 

[cookiesandcream]

to integrate of 5/3x, you need f'(x) on the top and f(x) on the bottom, so take 5 out of the picture:

5 S 1/3x dx (S meaning the integral sign)
= 5/3 S 3/3x dx(as you need a 3 on top, you borrow a three and then divide by it)
= 5/3 ln 3x + c
hope that helps

edit*lol i think i posted this a bit late

 

[micuzzo]

^yeah great stuff guys i now understand...thanks heaps esp to lolokay