MAth DOubts

[Twinkle Charms]

Find the derivatives of
(a) ln(1-2x)
(b) ln(a+bx)
(c) ln(x^2(x-1))
(d) ln(2x)

please post the answers asap i did it buh want to check , a bit confused, Thank You.

[Twinkle Charms]

wow 10 views ns till no reply?? =(
i got math exam ppl, need help...

[Kim]

wait ill ask a friend she might know she takes maths

[CHEMMASTER6000]

its probably cause of the implication of derivative. what do you mean dirivative of the ln functions. i dont think we need that for p3

[Kim]

my friend says the questions dont make sense lol

[Alpha]

d/dx [ln f(x)] = f '(x)/ f(x), where f ' (x) is the derivative of f(x).


Derivatives:

(a) ln(1-2x) > -2/ (1-2x)

(b) ln(a+bx) > b/ (a+bx)

(c) ln(x^2(x-1)) >>> clear this one, power is only 2 or 2(x-1)?

  I think you forgot to add the brackets, must have been  ln((x^2)(x-1)).
For this one, use the properties of logarithms. ln ((x^2)(x-1)) = ln (x^2) + ln (x-1)
                                                                       
Differentiate by step to get the derivative.

ln (x^2) > 2x/ (x^2)
ln (x-1) > 1/(x-1)

Therefore,  ln (x^2) + ln (x-1) = 2x/ (x^2) + 1/(x-1)

Simplify further if you want.

(d) ln(2x) > 2/2x

[Twinkle Charms]

d/dx [ln f(x)] = f '(x)/ f(x), where f ' (x) is the derivative of f(x).


Derivatives:

(a) ln(1-2x) > -2/ (1-2x)

(b) ln(a+bx) > b/ (a+bx)

(c) ln(x^2(x-1)) >>> clear this one, power is only 2 or 2(x-1)?

  I think you forgot to add the brackets, must have been  ln((x^2)(x-1)).
For this one, use the properties of logarithms. ln ((x^2)(x-1)) = ln (x^2) + ln (x-1)
                                                                       
Differentiate by step to get the derivative.

ln (x^2) > 2x/ (x^2)
ln (x-1) > 1/(x-1)

Therefore,  ln (x^2) + ln (x-1) = 2x/ (x^2) + 1/(x-1)

Simplify further if you want.

(d) ln(2x) > 2/2x

thankyou alpha, n
its x square multiplied by x-1, so for it we open the brackets n then differientiate? N how to do the diff. part?

[Twinkle Charms]

its probably cause of the implication of derivative. what do you mean dirivative of the ln functions. i dont think we need that for p3

yes ofc its there in p3, the main thing dude :p

[CHEMMASTER6000]

oh woops haha , my mistake  i didnt take derivative as differentiate. makes a load of sense now, i can explain here if the you dont understand the guy above

[Twinkle Charms]

oh woops haha , my mistake  i didnt take derivative as differentiate. makes a load of sense now, i can explain here if the you dont understand the guy above

nop i got her, actually i did it just wanted to confirm, cz sum1 was confusing me :|

[Twinkle Charms]

oh and also i find vectors difficult, if someone can gimme good notes or site from which to refer, Vectors in 2D n 3D? the vector equation n all tht crap :|

[Alpha]

thankyou alpha, n
its x square multiplied by x-1, so for it we open the brackets n then differientiate? N how to do the diff. part?

ln [x^2(x-1)] = ln [x^(2x-2)]

Using laws of indices,

                  = ln (x^2x/ x^2)

Using laws of logarithms,

                 = ln (x^2x) - ln (x^2)

Derivative > 2x. x^(2x-1) {which can be simplified to 2x^(2x-1+1)= 2x^2x}/ x^(2x) - 2x/ x^2

              = 2x^2x/ x^2x - 2x/ x^2

This one's more complex.
Welcome~

[Twinkle Charms]

ln [x^2(x-1)] = ln [x^(2x-2)]

Using laws of indices,

                  = ln (x^2x/ x^2)

Using laws of logarithms,

                 = ln (x^2x) - ln (x^2)

Derivative > 2x. x^(2x-1) {which can be simplified to 2x^(2x-1+1)= 2x^2x}/ x^(2x) - 2x/ x^2

              = 2x^2x/ x^2x - 2x/ x^2

This one's more complex.
Welcome~

uh-oh seems u neva got wht actually im asking :|

its x(square) this multiplied to (x-1) now get it??

[Alpha]

uh-oh seems u neva got wht actually im asking :|

its x(square) this multiplied to (x-1) now get it??

Yeah, but isn't that what I did earlier? My first post?

[Twinkle Charms]

Yeah, but isn't that what I did earlier? My first post?

ooopps my bad, i was reading it rong ur first post...sooorrryyy and thank youu so mucchh