Question

y=1/log5x find y'

Answer 1

is this
\[y=\frac{1}{\log(5x)}\] or perhaps
\[y=\frac{1}{\log_5(x)}\]?

Answer 2

second one

Answer 3

ok then use the following
\[(\frac{1}{f})'=-\frac{f'}{f^2}\] with
\[f(x)=\log_5(x),f'(x)=\frac{1}{x\ln(5)}\]

Answer 4

or if you prefer, recall from the change of base formula that
\[\log_5(x)=\frac{\ln(x)}{\ln(5)}\] and so
\[\frac{1}{\log_5(x)}=\frac{\ln(5)}{\ln(x)}\] and take the derivative of that . it amounts to the same thing

Answer 5

good from there or you need me to work it out?

Answer 6

that's a derivative rule or is that just you putting it simple?

Answer 7

Yes please. I forgot there was a 4 at the top.

Answer 8

the 4 is of no consequence. we just multiply by 4 at the end

Answer 9

we want
\[\frac{d}{dx}\frac{4\ln(5)}{\ln(x)}\] so we get
\[\frac{-4\ln(5)}{\ln^2(x)}\times \frac{1}{x}\]

Answer 10

@outcast i am not sure if your question is directed at my statement that
\[(\frac{1}{f})'=\frac{f'}{f^2}\] but that is easy to prove directly from the definition of the derivative. it is also a simple consequence of the chain rule because the derivative of
\[\frac{1}{x}\text { is } -\frac{1}{x^2}\] so by the chain rule the derivative of
\[\frac{1}{f(x)}\text { is } -\frac{1}{f^2(x)}\times f'(x)\]

Answer 11

missed a minus sign on first line sorry

Answer 12

Ah, yeah it's chain rule with either quotient or exponent raised to numerator. I thought maybe in another book, this was another derivative rule shown in the books. Idk why they don't cause after a while you'll eventually grasped the concept that when you have 1/f it will be what is above and then the nex will be the n!

Answer 13

not only that but when i took calc 1 i never saw it like that until i got to calc 2 and did sums

Answer 14

series* i mean in which you see it clearly

Answer 15

in fact it is much easier to start with that and then get the quotient rule. you can go directly from the definition and write
\[\lim_{h\rightarrow 0}\frac{\frac{1}{f(x+h)}-\frac{1}{f(x)}}{h}\]
\[\lim_{x\rightarrow 0}\frac{f(x)-f(x+h)}{hf(x)f(x+h)}\]
\[\lim_{h\rightarrow 0}\frac{f(x)-f(x+h)}{h}\times \lim_{h\rightarrow 0}\frac{1}{f(x)f(x+h)}\]
\[=-f'(x)\times \frac{1}{f^2(x)}\] assuming f is differentiable and therefore continuous etc

Answer 16

then get the quotient rule directly from that one using product rule, and without all the usual nonsense

Answer 17

so i guess what i am saying is that yes, this is done in some books first as a rule