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

Question

watchmath · Answer

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

anonymous · Answer

lol

amistre64 · Answer

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

mattfeury · Answer

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

anonymous · Answer

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

amistre64 · Answer

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 · Answer

DOH!!

anonymous · Answer

. is dot product here?

amistre64 · Answer

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

amistre64 · Answer

yeah '.' is lazt multiplication

amistre64 · Answer

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

watchmath · Answer

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

amistre64 · Answer

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 · Answer

its 64 watchmath...lol

amistre64 · Answer

itll be 65 after this; i think im aging lol

watchmath · Answer

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

anonymous · Answer

:-D

amistre64 · Answer

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 · Answer

just so you know amistre, the answer is zero

watchmath · Answer

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

amistre64 · Answer

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 · Answer

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

amistre64 · Answer

thats all im doin for it by the way lol

amistre64 · Answer

and its prolly chock full of typos :_)

shadowfiend · Answer

BOOOOOM! 600 MEDALER!

mattfeury · Answer

*pops champagne*

amistre64 · Answer

Mormons dont drink champagne....

mattfeury · Answer

*pops sparkling cider*

amistre64 · Answer

lol

shadowfiend · Answer

But everyone enjoys the nice popping sound of a cork!

anonymous · Answer

oh he opened it for himself

anonymous · Answer

one red punch for amistre and some sugar cookies

amistre64 · Answer

and orange jello with carrots....

shadowfiend · Answer

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

amistre64 · Answer

statistically; im an enigma ;)

watchmath · Answer

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

anonymous · Answer

$$x ^{m}.y ^{n}=(x+y)^{m+n}$$
so,y'=y/x
and y''=0

this is how its solved in my book. does anyone understand what they did in here?

watchmath · Answer

Now I like the problem :D.

anonymous · Answer

did you get it watchmath?

watchmath · Answer

No, not yet.

anonymous · Answer

ok...thats everything the solution says.

amistre64 · Answer

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

amistre64 · Answer

multiplying unlike bases...

watchmath · Answer

I got $y'=-y/x$ :)

watchmath · Answer

Is it y/x or -y/x ?

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

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

watchmath · Answer

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 · Answer

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

anonymous · Answer

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

watchmath · Answer

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 · Answer

oh.. i c now