Mathematics
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OpenStudy (anonymous):
for the equation given below, y/x+8y=x^9-4 evaluate dy/dx at point (1,-3/25)
12 years ago
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OpenStudy (freckles):
Troubles differentiating I suppose?
12 years ago
OpenStudy (freckles):
\[\frac{y}{x}+8y=x^9-4 \text{ is correct interpretation ?} \]
12 years ago
OpenStudy (anonymous):
sorry i meant (y)/(x+8y)=x^9-4
12 years ago
OpenStudy (anonymous):
8y under the y as well
12 years ago
OpenStudy (freckles):
\[\frac{y}{x+8y}=x^9-4\]
I bet you have no problem with the right hand side
12 years ago
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OpenStudy (freckles):
The left hand side will require us to know quotient rule
12 years ago
OpenStudy (freckles):
\[\frac{d}{dx}(\frac{y}{x+8y})=\frac{(y)'(x+8y)-y(x+8y)'}{(x+8y)^2}\]
12 years ago
OpenStudy (freckles):
This is only leaves you with differentiating y and (x+8y)
12 years ago
OpenStudy (freckles):
\[\frac{d}{dx}y=\frac{dy}{dx} \text{ (note: which people also call this) } =y'\]
12 years ago
OpenStudy (freckles):
Try finding
\[\frac{d}{dx}(x+8y)=?\]
12 years ago
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OpenStudy (anonymous):
ok i got,
[1+8 d/dx (y(x))\]
12 years ago
OpenStudy (freckles):
ok so we have 1+8y' coolness
12 years ago
OpenStudy (freckles):
ok so let's put all of that in...
12 years ago
OpenStudy (freckles):
\[\frac{y'(x+8y)-y(1+8y')}{(x+8y)^2}=9x^8\]
12 years ago
OpenStudy (freckles):
Now we need to solve for y'...
12 years ago
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OpenStudy (freckles):
first I would multiply both side by that ickyness in the bottom on the left side
12 years ago
OpenStudy (freckles):
Then after you have done
I would start using distributive property where needed on left.
Then gather my terms that y' and put everything else on the other side
12 years ago
OpenStudy (anonymous):
i dont get what you mean by multiply both sides
12 years ago
OpenStudy (anonymous):
multiply both sides by what?
12 years ago
OpenStudy (freckles):
by the ickyness on the bottom on the left
12 years ago
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OpenStudy (freckles):
There is a fraction on the left
a fraction has a top and a bottom
12 years ago
OpenStudy (freckles):
\[\cancel{(x+8y)^2} \cdot \frac{y'(x+8y)-y(1+8y')}{\cancel{(x+8y)^2}}=9x^8(x+8y)^2\]
12 years ago
OpenStudy (freckles):
I multiplied both sides by the ickyness that was on the bottom of the left
12 years ago
OpenStudy (freckles):
I'm doing this because I'm trying to isolate y'
12 years ago
OpenStudy (freckles):
That is the side with y'
12 years ago
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OpenStudy (freckles):
To get to y' I have to undo that division first
12 years ago
OpenStudy (anonymous):
oh ok..
12 years ago
OpenStudy (anonymous):
got you
12 years ago
OpenStudy (anonymous):
now do i distribute on the left?
12 years ago
OpenStudy (freckles):
Yep
If you can actually skip that step if you are really good with algebra and sorta like basically do it in your head
but yeah
12 years ago
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OpenStudy (freckles):
You want to gather your terms that have y' together
12 years ago
OpenStudy (anonymous):
xy'+8yy'-7+8yy'?
12 years ago
OpenStudy (anonymous):
is that correct for the left side?
12 years ago
OpenStudy (freckles):
\[xy'+8yy'-y-8yy'\]
12 years ago
OpenStudy (anonymous):
yah i meant y instead of 7 lol
12 years ago
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OpenStudy (freckles):
and a - in front of the 8yy'
12 years ago
OpenStudy (freckles):
but anyways you should see something cancel there
12 years ago
OpenStudy (freckles):
before we move on
12 years ago
OpenStudy (anonymous):
so xy'-y
12 years ago
OpenStudy (freckles):
\[xy'-y=9x^8(x+8y)^2\]
12 years ago
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OpenStudy (freckles):
Do you think you can finish solving for y' now?
12 years ago
OpenStudy (freckles):
You only need 2 more steps
12 years ago
OpenStudy (anonymous):
add y and divide by x correct?
12 years ago
OpenStudy (freckles):
yep
12 years ago
OpenStudy (freckles):
\[y'=\frac{9x^8(x+8y)^2+y}{x}\]
12 years ago
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OpenStudy (anonymous):
yep now i will plug in the x and y values
12 years ago
OpenStudy (anonymous):
and see if its right
12 years ago
OpenStudy (freckles):
yeah
12 years ago
OpenStudy (anonymous):
ok one sec
12 years ago
OpenStudy (anonymous):
-1.056
12 years ago
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OpenStudy (anonymous):
and it is correct
12 years ago
OpenStudy (freckles):
Well I will have to check again
I got -.1056
maybe i made mistake then
12 years ago
OpenStudy (anonymous):
thank you freckles! can you help me with another question
12 years ago
OpenStudy (anonymous):
no it is right, i submitted it and it came up correct
12 years ago
OpenStudy (freckles):
oh I thought you put in -1.056
12 years ago
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OpenStudy (anonymous):
oh no .-1056
12 years ago
OpenStudy (freckles):
ok cool
12 years ago
OpenStudy (anonymous):
suppose that (g(x))^2+6x=x^2g(x)+17 and g(2) =5 find g'(2)
12 years ago