LilySwan
find dy/dx if
(ax+cx^5)/kx^8
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myininaya
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a,c, and k are constants?
LilySwan
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yes
micahwood50
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Just use quotient rule.
LilySwan
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got stucked at one point
micahwood50
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\(\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\)
LilySwan
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why can we solve the numerator part first
micahwood50
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treat constant as if they are just "number", and not variable.
myininaya
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@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?
LilySwan
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for the upper part 5cx^4 +a
micahwood50
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Right f'(x) = a+5cx^4
micahwood50
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Can you find g'(x)?
LilySwan
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for the lower part it will be 8kx^7
micahwood50
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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.
micahwood50
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Can you do it? @LilySwan
LilySwan
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trying wait
LilySwan
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i was told to use this formula
\[\frac{ d }{ dx }(\frac{ u}{ v }) =\frac{ (v \frac{ du }{ dx })-(u \frac{ dv }{ dx }) }{ v^2 }\]
myininaya
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same formula
different notation
LilySwan
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?
myininaya
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\[\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
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
LilySwan
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\[5cx^4+a.kx^8 - ax+cx^5. 8kx^7/(kx^8)^2\]
myininaya
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\[\frac{(5cx^4+a)(kx^8)-(ax+cx^5)(8kx^7)}{(kx^8)^2}\]
i think this is what you mean
LilySwan
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yes
LilySwan
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what should i be doing the next
myininaya
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guess you could multiply things out on top
and also use law of exponents on the bottom to simplify the bottom