Differentiation of loagrithms (1 Viewer)

[micuzzo]

hi can someone explain this question for me..??

differentiate 1-loge 3x (i think it can also be written as 1-In3x)
answer= -1/x ... i thought it would be -1/3x

any ideas???

thanking you..

 

[Pwnage101]

hi can someone explain this question for me..??

differentiate 1-loge 3x (i think it can also be written as 1-In3x)
answer= -1/x ... i thought it would be -1/3x

any ideas???

thanking you..

number 1 , the short had for Logex is lnx (with an L not an I)

number 2, to differentiate a logarithm to base e, remember it is 'the derivative over the function', so if in the form of y = ln [f(x)], the derivative is y' = f'(x) / f(x)

so the derivative of 1-ln3x is = -3/3x = -1/x

 

[lyounamu]

hi can someone explain this question for me..??

differentiate 1-loge 3x (i think it can also be written as 1-In3x)
answer= -1/x ... i thought it would be -1/3x

any ideas???

thanking you..

d/dx (lnf(x)) = f'(x)/f(x)

so d/dx (1-ln(3x)) = -3/3x = -1/x

 

[micuzzo]

oh ok... i see where i went wrong... thank you

 

[micuzzo]

okay ive got another 1>>>

ln[root(2-x)], find f'(1)

my working: 0.5(2-x)^-0.5/root(2-x)

when subbing 1 i get 0.5 (or maybe its meant to be +/- 0.5) however answer is -0.5...

can some one please help... im beggining to think that the given answer is wrong!

EDIT: okay now these things are getting annoying, ive got another 1... differentiate log x ( i mean log to the base of 10)

i used the base rules so i get lnx/ln10 and then used the quotient rule... but in doing this i will get zero... i saw an example but dont understand it, it goes lik this: diff: log (base 2) x
base rules= lnx/ln2
= 1/ln2 times lnx
d/dx = 1/ln2 times 1/x <<<< i dont understand this line... wot happens to the 1/ln2 dont we differntiate it
= 1/xln2

sorry for my confusing format

 

[Timothy.Siu]

okay ive got another 1>>>

ln[root(2-x)], find f'(1)

my working: 0.5(2-x)^-0.5/root(2-x)

when subbing 1 i get 0.5 (or maybe its meant to be +/- 0.5) however answer is -0.5...

can some one please help... im beggining to think that the given answer is wrong!

u differentiated it wrong,

d/dx root(2-x)=-0.5(2-x)^-0.5

 

[lyounamu]

okay ive got another 1>>>

ln[root(2-x)], find f'(1)

my working: 0.5(2-x)^-0.5/root(2-x)

when subbing 1 i get 0.5 (or maybe its meant to be +/- 0.5) however answer is -0.5...

can some one please help... im beggining to think that the given answer is wrong!

root(2-x) = (2-x)^1/2

derivative is 1/2 (2-x)^-0.5 x - 1 = -1/(2(2-x)^1/2)

f'(ln(root(2-x)) = -1/(2(2-x)^1/2)/root(2-x)
f'(1) = -1/2

 

[micuzzo]

root(2-x) = (2-x)^1/2

derivative is 1/2 (2-x)^-0.5 x - 1 = -1/(2(2-x)^1/2)

f'(ln(root(2-x)) = -1/(2(2-x)^1/2)/root(2-x)
f'(1) = -1/2

AHHHHH stupid syntax errors!!!!!!!!!!!!! ic function of function

thanks

 

[micuzzo]

DIT: okay now these things are getting annoying, ive got another 1... differentiate log x ( i mean log to the base of 10)

i used the base rules so i get lnx/ln10 and then used the quotient rule... but in doing this i will get zero... i saw an example but dont understand it, it goes lik this: diff: log (base 2) x
base rules= lnx/ln2
= 1/ln2 times lnx
d/dx = 1/ln2 times 1/x <<<< i dont understand this line... wot happens to the 1/ln2 dont we differntiate it
= 1/xln2

sorry for my confusing format can u please help me with this 1

 

[lyounamu]

DIT: okay now these things are getting annoying, ive got another 1... differentiate log x ( i mean log to the base of 10)

i used the base rules so i get lnx/ln10 and then used the quotient rule... but in doing this i will get zero... i saw an example but dont understand it, it goes lik this: diff: log (base 2) x
base rules= lnx/ln2
= 1/ln2 times lnx
d/dx = 1/ln2 times 1/x <<<< i dont understand this line... wot happens to the 1/ln2 dont we differntiate it
= 1/xln2

sorry for my confusing format can u please help me with this 1

\

f(x) = log(2) x = ln x/ln2
f'(x) = 1/xln2

 

[micuzzo]

wot about ln2... dont we differentiate

 

[lyounamu]

wot about ln2... dont we differentiate

tell me how you differentiate 2x

isn't that just 2?

this is the same case here, ln2 is just a coefficient.

 

[clintmyster]

I sense another question coming haha

 

[lyounamu]

I sense another question coming haha

I think he is typing it. Pity that I won't be able to do it because I need to retreat to my bed sooner or later...

 

[micuzzo]

hang on a second.... can you re do the question with the working out
the question is: y=log(2) x
answer= 1/xlog(e) 2

sorry... still a bit confused

 

[micuzzo]

I sense another question coming haha

lol

 

[lyounamu]

hang on a second.... can you re do the question with the working out
the question is: y=log(2) x
answer= 1/xlog(e) 2

sorry... still a bit confused

oh sorry, i left the x part

it's 1/xln2

i left x

haha

because if you diff lnx, that's 1/x and 1/ln2 stays behind.

 

[micuzzo]

i thought something fishy was going on when i gave u the answer with the post lol.... but now u left out the working...ahahah

 

[cutemouse]

hang on a second.... can you re do the question with the working out
the question is: y=log(2) x
answer= 1/xlog(e) 2

sorry... still a bit confused

log₂x=log_ex/log_e2 [Log rules of changing base]

By quotient rule, d/dx=(vu'-uv')/v²

Therefore, d/dx=[(log_e2).(1/x)-log_ex.0]/(log_e2)²

=1/xlog_e2
=1/xln2

Hope that helps

 

[jet]

Its that the ln2 is an actual number. When you enter that into your calculator you get an actual number. Try it. so, d/dx(lnx/ln2)=(1/ln2)d/dx(lnx)
=1/xln2

Alot of people are confused by that. Normally, if your calculator can spit out an answer (not 'Error 2' lol) then you dont count it in the differentiation.