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MrMoose Group TitleBest ResponseYou've already chosen the best response.0
this involves both the chain rule and the quotient rule
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
or you can use log laws to change it first; then diferentiate it, would be easier.
 2 years ago

cunninnc Group TitleBest ResponseYou've already chosen the best response.0
@failmathmajor only the 2/x^3 is a fraction
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
is it \[\log\left(5x^{4}\frac{2}{x^{3}}\right) ?\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
that would be much easier
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
so using log properties you can say that \[\log_{10} {x} = \ln x / \ln 10 \]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
ln10 is a constant, you can factor that out
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
can you use the chain rule fluently?
 2 years ago

cunninnc Group TitleBest ResponseYou've already chosen the best response.0
y'=pu(x) * u'(x) right?
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
d(f(g(x))/dx = (dg(x)/dx)*(df(g(x))/dx) is the way I know it
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
Or..\[y= \log\left(5x^{4}\frac{2}{x^{3}}\right) y = \log\left(\frac{5x^{7}2}{x^{3}}\right) => y = \log\left(5x^{7}2\right)  \log(x^3)\] Might be easier; than the quotient role and chain rule..
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
that may look confusing now that I see it
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
don't need the quotient in this case: can treat /x^3 as a x^3
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
@cunninnc: are you able to do it now?
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
and you would still need the chain rule for that anyway
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
but enough arguing
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
Did an argument started? lol
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
\[(d(\ln (5x^4  2x^{3}))/dx )/ \ln10\] is what it simplifies down to, to be concise
 2 years ago

cunninnc Group TitleBest ResponseYou've already chosen the best response.0
@Mimi_x3 kinda .... i see mr. moose got 2x^3 where does ^3 come froms
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
\[2/x^3= 2 * x^{3}\]
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
Sorry i don't know what MrMoose is doing. @MrMoose: Use \frac{x}{y} for fractions :)
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
Or why not try the method that i used :) \[ \frac{d}{dx} \log\left(5x^{7}2\right) \frac{d}{dx} \log(x^3)\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
when you divide you subtract exponents, so that is equivalent to saying \[2 * \frac{x^0}{x^3}\] then subtract exponents in division
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{\frac{d(\ln(5x^4−2x^{−3}))}{dx}}{\ln10}\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
@catamountz15 what?
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
I am almost entirely sure that that isn't a form of the answer.
 2 years ago

catamountz15 Group TitleBest ResponseYou've already chosen the best response.0
Here are the steps in to solving this problem.
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.0
that isn't what you wrote though
 2 years ago

catamountz15 Group TitleBest ResponseYou've already chosen the best response.0
My apologies that was an answer to a different problem.
 2 years ago
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