anonymous
  • anonymous
If x^6. y^2=(x+y)^8, then y''(double derivative) is equal to
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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watchmath
  • watchmath
I don't like this .... Go ahead amistre64 :D
anonymous
  • anonymous
lol
amistre64
  • amistre64
6x^5.y^2 + x^6.2y.y' = 7(x+y)^7 . y'

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mattfeury
  • mattfeury
(for medal #600. no pressure amistre :). )
anonymous
  • anonymous
need an 8 in front of that (x+y)^(7) term
amistre64
  • amistre64
6x^5.y^2 = 7(x+y)^7 . y' - x^6.2y.y' 6x^5.y^2 = (7(x+y)^7 - x^6.2y) y' 6x^5.y^2 ----------------- = y' ; whew!! first is down lol (7(x+y)^7 - x^6.2y)
amistre64
  • amistre64
DOH!!
anonymous
  • anonymous
. is dot product here?
amistre64
  • amistre64
6x^5.y^2 ----------------- = y' ; there lol (8(x+y)^7 - x^6.2y)
amistre64
  • amistre64
yeah '.' is lazt multiplication
amistre64
  • amistre64
y' = (6x^5) (y^2) (7(x+y)^7 - x^6.2y)^-1 to make life easier right?
watchmath
  • watchmath
Go Amistre65 for the 600 medal!! many people are watching :).
amistre64
  • amistre64
y'' = f'gh + fg'h + fgh' y'' = (6x^5) (y^2) (8(x+y)^7 - x^6.2y)^-1 <-- and i fixed the typo
anonymous
  • anonymous
its 64 watchmath...lol
amistre64
  • amistre64
itll be 65 after this; i think im aging lol
watchmath
  • watchmath
after you give him one more medal his username will change to 65 :D
anonymous
  • anonymous
:-D
amistre64
  • amistre64
y'' = (30x^4) (y^2) (8(x+y)^7 - x^6.2y)^-1 + (6x^5) (2y)y' (8(x+y)^7 - x^6.2y)^-1 -(6x^5) (y^2) (8(x+y)^7 - x^6.2y )^-2 (innards)
anonymous
  • anonymous
just so you know amistre, the answer is zero
watchmath
  • watchmath
Oh, you shouldn't let him know right now. Tell him later after he finish :).
amistre64
  • amistre64
innards = 56(x+y)^6.y' - [(x^6 . 2.y') + (6x^5.2y)] --------------------------------------- y'' = (30x^4) (y^2) (8(x+y)^7 - x^6.2y)^-1 + (6x^5) (2y)y' (8(x+y)^7 - x^6.2y)^-1 -(6x^5) (y^2) (8(x+y)^7 - x^6.2y )^-2 [*] *[56(x+y)^6.y' - (x^6. 2.y') - (6x^5.2y) ] = 0 lol
amistre64
  • amistre64
fill in all the y' bits and go to town ;)
amistre64
  • amistre64
thats all im doin for it by the way lol
amistre64
  • amistre64
and its prolly chock full of typos :_)
shadowfiend
  • shadowfiend
BOOOOOM! 600 MEDALER!
mattfeury
  • mattfeury
*pops champagne*
amistre64
  • amistre64
Mormons dont drink champagne....
mattfeury
  • mattfeury
*pops sparkling cider*
amistre64
  • amistre64
lol
shadowfiend
  • shadowfiend
But everyone enjoys the nice popping sound of a cork!
anonymous
  • anonymous
oh he opened it for himself
anonymous
  • anonymous
one red punch for amistre and some sugar cookies
amistre64
  • amistre64
and orange jello with carrots....
shadowfiend
  • shadowfiend
400 fans, 600 medals. You add up to 1000 now!
amistre64
  • amistre64
statistically; im an enigma ;)
watchmath
  • watchmath
There should be a shorter way to do this. Anybody knows?
anonymous
  • anonymous
\[x ^{m}.y ^{n}=(x+y)^{m+n}\] so,y'=y/x and y''=0 this is how its solved in my book. does anyone understand what they did in here?
watchmath
  • watchmath
Now I like the problem :D.
anonymous
  • anonymous
did you get it watchmath?
watchmath
  • watchmath
No, not yet.
anonymous
  • anonymous
ok...thats everything the solution says.
amistre64
  • amistre64
that would mean theres a rule of exponents for adding unlike bases somewhere right?
amistre64
  • amistre64
multiplying unlike bases...
watchmath
  • watchmath
I got \(y'=-y/x\) :)
watchmath
  • watchmath
Is it y/x or -y/x ?
anonymous
  • anonymous
x ^{m}.y ^{n}=(x+y)^{m+n} this relation is not correct.. \[2^{2}.3^{2}=4*9=36\] RHS:\[(x+y)^{m+n}=(2+3)^{2+2}=5^{4}=625\]
watchmath
  • watchmath
Take natural log of both sides we have \[m\ln x+n\ln y=(m+n)\ln(x+y)\] Then \(\frac{m}{x}+\frac{ny'}{y}=\frac{(m+n)(1+y')}{x+y}\) \(\frac{m(x+y)}{x}+\frac{n(x+y)y'}{y}=\frac{(m+n)x}{x}+\frac{y(m+n)y'}{y}\) \(\frac{n(x+y)-(m+n)y}{y}\cdot y'=\frac{(m+n)x-m(x+y)}{x}\) \(\frac{nx-my}{y}\cdot y'=\frac{nx-my}{x}\) Hence \(y'=y/x\) It follows that \(xy'=y\) Then \(y'+xy''=y''\) \(xy''=0\) \(y''=0\) :D :D
anonymous
  • anonymous
how did you proceed from this step y'+xy''=y''
anonymous
  • anonymous
and this relation \[x ^{m}.y ^{n}=(x+y)^{m+n}\] is not true. u do agree with me right?
watchmath
  • watchmath
The relation doesn't need to be true for all \(x,y\). Like \(x^2+y^2=1\) doesn't need to hold true for all \(x,y\). From \(xy'=y\) take the derivative implicityly (of course we need to use the product rule on the left)
watchmath
  • watchmath
You have an interesting book. What is the title of your book?
anonymous
  • anonymous
actually its a solution sheet of a practice question papers i attempted. y'+xy''=y'' xy''=0 plz explain these two steps
watchmath
  • watchmath
There was a typo on my solution above \(xy'=y\) \(y'+xy''=y'\) \(xy''=0\) If \(x\neq 0\) then \(y''=0\) If \(x=0\), then plug in to the original equation we have \(y=0\) which implies that \(y''=0\) as well.
anonymous
  • anonymous
oh.. i c now

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