Question

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

Answer 1

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

Answer 2

lol

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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)

Answer 7

DOH!!

Answer 8

. is dot product here?

Answer 9

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

Answer 10

yeah '.' is lazt multiplication

Answer 11

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

Answer 12

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

Answer 13

y'' = f'gh + fg'h + fgh'

y'' = (6x^5) (y^2) (8(x+y)^7 - x^6.2y)^-1 <-- and i fixed the typo

Answer 14

its 64 watchmath...lol

Answer 15

itll be 65 after this; i think im aging lol

Answer 16

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

Answer 17

:-D

Answer 18

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)

Answer 19

just so you know amistre, the answer is zero

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

thats all im doin for it by the way lol

Answer 24

and its prolly chock full of typos :_)

Answer 25

BOOOOOM! 600 MEDALER!

Answer 26

*pops champagne*

Answer 27

Mormons dont drink champagne....

Answer 28

*pops sparkling cider*

Answer 29

lol

Answer 30

But everyone enjoys the nice popping sound of a cork!

Answer 31

oh he opened it for himself

Answer 32

one red punch for amistre and some sugar cookies

Answer 33

and orange jello with carrots....

Answer 34

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

Answer 35

statistically; im an enigma ;)

Answer 36

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

Answer 37

\[x ^{m}.y ^{n}=(x+y)^{m+n}\]
so,y'=y/x
and y''=0

this is how its solved in my book. does anyone understand what they did in here?

Answer 38

Now I like the problem :D.

Answer 39

did you get it watchmath?

Answer 40

No, not yet.

Answer 41

ok...thats everything the solution says.

Answer 42

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

Answer 43

multiplying unlike bases...

Answer 44

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

Answer 45

Is it y/x or -y/x ?

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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)

Answer 51

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

Answer 52

actually its a solution sheet of a practice question papers i attempted.


y'+xy''=y''
xy''=0
plz explain these two steps

Answer 53

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.

Answer 54

oh.. i c now