find the derivative of y=log10(5x^4-2/x^3)

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find the derivative of y=log10(5x^4-2/x^3)

Mathematics
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this involves both the chain rule and the quotient rule
or you can use log laws to change it first; then diferentiate it, would be easier.
@failmathmajor only the 2/x^3 is a fraction

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

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