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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

MrMoose
 2 years ago
Best 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

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

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

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

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

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

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

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

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

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