Quick Question

[IBSQUARED]

Could someone please go over differentiating y = ln (5x) using first principles


The answer should come out to 1/x

Thanks

[moneyfaery]

Do you mean 5/x?

The derivative of lnx is 1/x

[Hedron123]

As moneyfaery said: the derivative of ln x is equal to 1/x.
The derivative of ln (5x) is = 1 /5x times 5 (derivative of 5x).
So: the answer is 1 / x

[moneyfaery]

Is it? Oops, can't remember. It's been a year.

[IBSQUARED]

Yeah i know the ln rule.. but a question in our text asked to prove its 1/x using first principles

so the f(x+h) - f(x)
--------------
h

[Abu]

IBSQUARED - We don't do first principles for Maths HL

[ezex]

Cmon...give the poor kid a hand! Here goes, try to keep up:

deriv of ln(x) = lim(h->0) [ ln(x+h) - ln(x) ] / h from the definition of the deriv.
= lim ln((x+h)/x) / h from log rules when subtracting two logs
= lim (1/h) ln(1 + h/x) just took the 1/h outside
= lim [ ln (1 + h/x)^(1/h) ]. rules of logs when you convert the coeficient to exponent

Set u=h/x and substitute (so things fit , mathematicians get to do that):

lim(u->0) [ ln (1 + u)^(1/(ux)) ] replaced formula
= 1/x ln [ lim(u->0) (1 + u)^(1/u) ] brought down a 1/x from the exponent using rules of logs again
= 1/x ln (e) replaced that whole mess after ln by e because lim(u->0) (1 + u)^(1/u) is the formal definition of e, look it up
= 1/x. uh...ya, if you don't know what i did to get to this step you probably didn't follow on the proof so i'm not gonna say that ln (e) = 1...dammit

[Abu]

Not all of us are math prodigies!

[ezex]

pfft this was math was baby...lol