cunninnc
find the derivative of y=log10(5x^4-2/x^3)
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MrMoose
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this involves both the chain rule and the quotient rule
Mimi_x3
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or you can use log laws to change it first; then diferentiate it, would be easier.
cunninnc
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@failmathmajor only the 2/x^3 is a fraction
Mimi_x3
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is it
\[\log\left(5x^{4}-\frac{2}{x^{3}}\right) ?\]
MrMoose
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that would be much easier
MrMoose
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so using log properties you can say that \[\log_{10} {x} = \ln x / \ln 10 \]
MrMoose
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ln10 is a constant, you can factor that out
MrMoose
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can you use the chain rule fluently?
cunninnc
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y'=pu(x) * u'(x) right?
MrMoose
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d(f(g(x))/dx = (dg(x)/dx)*(df(g(x))/dx) is the way I know it
Mimi_x3
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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..
MrMoose
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that may look confusing now that I see it
MrMoose
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don't need the quotient in this case:
can treat /x^3 as a x^-3
Mimi_x3
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@cunninnc: are you able to do it now?
MrMoose
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and you would still need the chain rule for that anyway
MrMoose
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but enough arguing
Mimi_x3
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Did an argument started? lol
MrMoose
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\[(d(\ln (5x^4 - 2x^{-3}))/dx )/ \ln10\]
is what it simplifies down to, to be concise
cunninnc
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@Mimi_x3 kinda .... i see mr. moose got 2x^-3 where does ^-3 come froms
MrMoose
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\[2/x^3= 2 * x^{-3}\]
Mimi_x3
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Sorry i don't know what MrMoose is doing.
@MrMoose: Use \frac{x}{y} for fractions :)
Mimi_x3
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Or why not try the method that i used :)
\[ \frac{d}{dx} \log\left(5x^{7}-2\right) -\frac{d}{dx} \log(x^3)\]
MrMoose
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when you divide you subtract exponents, so that is equivalent to saying \[2 * \frac{x^0}{x^3}\]
then subtract exponents in division
MrMoose
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\[\frac{\frac{d(\ln(5x^4−2x^{−3}))}{dx}}{\ln10}\]
MrMoose
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@catamountz15
what?
MrMoose
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I am almost entirely sure that that isn't a form of the answer.
catamountz15
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Here are the steps in to solving this problem.
MrMoose
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that isn't what you wrote though
catamountz15
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My apologies that was an answer to a different problem.