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anonymous
 5 years ago
If x^6. y^2=(x+y)^8, then y''(double derivative) is equal to
anonymous
 5 years ago
If x^6. y^2=(x+y)^8, then y''(double derivative) is equal to

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I don't like this .... Go ahead amistre64 :D

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.56x^5.y^2 + x^6.2y.y' = 7(x+y)^7 . y'

mattfeury
 5 years ago
Best ResponseYou've already chosen the best response.0(for medal #600. no pressure amistre :). )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0need an 8 in front of that (x+y)^(7) term

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.56x^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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0. is dot product here?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.56x^5.y^2  = y' ; there lol (8(x+y)^7  x^6.2y)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5yeah '.' is lazt multiplication

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5y' = (6x^5) (y^2) (7(x+y)^7  x^6.2y)^1 to make life easier right?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Go Amistre65 for the 600 medal!! many people are watching :).

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5y'' = f'gh + fg'h + fgh' y'' = (6x^5) (y^2) (8(x+y)^7  x^6.2y)^1 < and i fixed the typo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its 64 watchmath...lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5itll be 65 after this; i think im aging lol

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0after you give him one more medal his username will change to 65 :D

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5y'' = (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
 5 years ago
Best ResponseYou've already chosen the best response.0just so you know amistre, the answer is zero

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, you shouldn't let him know right now. Tell him later after he finish :).

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5innards = 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
 5 years ago
Best ResponseYou've already chosen the best response.5fill in all the y' bits and go to town ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5thats all im doin for it by the way lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5and its prolly chock full of typos :_)

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0BOOOOOM! 600 MEDALER!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5Mormons dont drink champagne....

mattfeury
 5 years ago
Best ResponseYou've already chosen the best response.0*pops sparkling cider*

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0But everyone enjoys the nice popping sound of a cork!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh he opened it for himself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one red punch for amistre and some sugar cookies

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5and orange jello with carrots....

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0400 fans, 600 medals. You add up to 1000 now!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5statistically; im an enigma ;)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0There should be a shorter way to do this. Anybody knows?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x ^{m}.y ^{n}=(x+y)^{m+n}\] so,y'=y/x and y''=0 this is how its solved in my book. does anyone understand what they did in here?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Now I like the problem :D.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you get it watchmath?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...thats everything the solution says.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5that would mean theres a rule of exponents for adding unlike bases somewhere right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.5multiplying unlike bases...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x ^{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
 5 years ago
Best ResponseYou've already chosen the best response.0Take natural log of both sides we have \[m\ln x+n\ln y=(m+n)\ln(x+y)\] Then \(\frac{m}{x}+\frac{ny'}{y}=\frac{(m+n)(1+y')}{x+y}\) \(\frac{m(x+y)}{x}+\frac{n(x+y)y'}{y}=\frac{(m+n)x}{x}+\frac{y(m+n)y'}{y}\) \(\frac{n(x+y)(m+n)y}{y}\cdot y'=\frac{(m+n)xm(x+y)}{x}\) \(\frac{nxmy}{y}\cdot y'=\frac{nxmy}{x}\) Hence \(y'=y/x\) It follows that \(xy'=y\) Then \(y'+xy''=y''\) \(xy''=0\) \(y''=0\) :D :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you proceed from this step y'+xy''=y''

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and this relation \[x ^{m}.y ^{n}=(x+y)^{m+n}\] is not true. u do agree with me right?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0You have an interesting book. What is the title of your book?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually its a solution sheet of a practice question papers i attempted. y'+xy''=y'' xy''=0 plz explain these two steps

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0There 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.
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