Ask your own question, for FREE!
Mathematics 63 Online
OpenStudy (anonymous):

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

OpenStudy (watchmath):

I don't like this .... Go ahead amistre64 :D

OpenStudy (anonymous):

lol

OpenStudy (amistre64):

6x^5.y^2 + x^6.2y.y' = 7(x+y)^7 . y'

OpenStudy (mattfeury):

(for medal #600. no pressure amistre :). )

OpenStudy (anonymous):

need an 8 in front of that (x+y)^(7) term

OpenStudy (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)

OpenStudy (amistre64):

DOH!!

OpenStudy (anonymous):

. is dot product here?

OpenStudy (amistre64):

6x^5.y^2 ----------------- = y' ; there lol (8(x+y)^7 - x^6.2y)

OpenStudy (amistre64):

yeah '.' is lazt multiplication

OpenStudy (amistre64):

y' = (6x^5) (y^2) (7(x+y)^7 - x^6.2y)^-1 to make life easier right?

OpenStudy (watchmath):

Go Amistre65 for the 600 medal!! many people are watching :).

OpenStudy (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

OpenStudy (anonymous):

its 64 watchmath...lol

OpenStudy (amistre64):

itll be 65 after this; i think im aging lol

OpenStudy (watchmath):

after you give him one more medal his username will change to 65 :D

OpenStudy (anonymous):

:-D

OpenStudy (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)

OpenStudy (anonymous):

just so you know amistre, the answer is zero

OpenStudy (watchmath):

Oh, you shouldn't let him know right now. Tell him later after he finish :).

OpenStudy (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

OpenStudy (amistre64):

fill in all the y' bits and go to town ;)

OpenStudy (amistre64):

thats all im doin for it by the way lol

OpenStudy (amistre64):

and its prolly chock full of typos :_)

OpenStudy (shadowfiend):

BOOOOOM! 600 MEDALER!

OpenStudy (mattfeury):

*pops champagne*

OpenStudy (amistre64):

Mormons dont drink champagne....

OpenStudy (mattfeury):

*pops sparkling cider*

OpenStudy (amistre64):

lol

OpenStudy (shadowfiend):

But everyone enjoys the nice popping sound of a cork!

OpenStudy (anonymous):

oh he opened it for himself

OpenStudy (anonymous):

one red punch for amistre and some sugar cookies

OpenStudy (amistre64):

and orange jello with carrots....

OpenStudy (shadowfiend):

400 fans, 600 medals. You add up to 1000 now!

OpenStudy (amistre64):

statistically; im an enigma ;)

OpenStudy (watchmath):

There should be a shorter way to do this. Anybody knows?

OpenStudy (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?

OpenStudy (watchmath):

Now I like the problem :D.

OpenStudy (anonymous):

did you get it watchmath?

OpenStudy (watchmath):

No, not yet.

OpenStudy (anonymous):

ok...thats everything the solution says.

OpenStudy (amistre64):

that would mean theres a rule of exponents for adding unlike bases somewhere right?

OpenStudy (amistre64):

multiplying unlike bases...

OpenStudy (watchmath):

I got \(y'=-y/x\) :)

OpenStudy (watchmath):

Is it y/x or -y/x ?

OpenStudy (anonymous):

x ^{m}.y ^{n}=(x+y)^{m+n} this relation is not correct.. \[2^{2}.3^{2}=4*9=36\] RHS:\[(x+y)^{m+n}=(2+3)^{2+2}=5^{4}=625\]

OpenStudy (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

OpenStudy (anonymous):

how did you proceed from this step y'+xy''=y''

OpenStudy (anonymous):

and this relation \[x ^{m}.y ^{n}=(x+y)^{m+n}\] is not true. u do agree with me right?

OpenStudy (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)

OpenStudy (watchmath):

You have an interesting book. What is the title of your book?

OpenStudy (anonymous):

actually its a solution sheet of a practice question papers i attempted. y'+xy''=y'' xy''=0 plz explain these two steps

OpenStudy (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.

OpenStudy (anonymous):

oh.. i c now

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
MAGABACK: ART!
2 hours ago 5 Replies 0 Medals
danielfootball123: Is Donald trump a good president?
3 hours ago 92 Replies 6 Medals
Gucchi: chem help
15 hours ago 9 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!