find dy/dx if (ax+cx^5)/kx^8

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find dy/dx if (ax+cx^5)/kx^8

Mathematics
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a,c, and k are constants?
yes
Just use quotient rule.

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got stucked at one point
\(\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\)
why can we solve the numerator part first
treat constant as if they are just "number", and not variable.
@LilySwan Did you have a problem with finding derivative of the top or the bottom? That is what you need to plug into @micahwood50 's formula. what do you mean solve the numerator? You shouldn't be solving. You should be differentiating it. What is the derivative of the top part of your fraction?
for the upper part 5cx^4 +a
Right f'(x) = a+5cx^4
Can you find g'(x)?
for the lower part it will be 8kx^7
Yep. So you have f(x) = ax+cx^5, f'(x) = a+5cx^4, g(x) = kx^8, g'(x) = 8kx^7 Just plug these in formula given above then simplify.
Can you do it? @LilySwan
trying wait
i was told to use this formula \[\frac{ d }{ dx }(\frac{ u}{ v }) =\frac{ (v \frac{ du }{ dx })-(u \frac{ dv }{ dx }) }{ v^2 }\]
same formula different notation
?
\[\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] if you want you can replace the f's with u's the g's with v's the ' things with d/dx and you have the same formula
\[5cx^4+a.kx^8 - ax+cx^5. 8kx^7/(kx^8)^2\]
\[\frac{(5cx^4+a)(kx^8)-(ax+cx^5)(8kx^7)}{(kx^8)^2}\] i think this is what you mean
yes
what should i be doing the next
guess you could multiply things out on top and also use law of exponents on the bottom to simplify the bottom

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