anonymous
  • anonymous
find second derivative if (x-5)^5+(y-2)^5=64
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
f1{x,y} = 5(x-5)^4 f11{x,y} = 20(x-5)^3
amistre64
  • amistre64
f2{x,y} = 5(y-2)^4 f22{x,y} = 20(y-2)^3
anonymous
  • anonymous
5(x-5)^4 + 5(y-2)^4 dy/dx =0 ist derivative

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amistre64
  • amistre64
right, so long as we imply that y is a function of x
anonymous
  • anonymous
yes,, maybe we can let snenlaha do the 2nd derivative hehehehe
amistre64
  • amistre64
maybe :)
anonymous
  • anonymous
i couldn't continue from the first to the 2nd derivative
anonymous
  • anonymous
but do you know the process of the ist that i did?
anonymous
  • anonymous
yes i do and i got it the same as yours
anonymous
  • anonymous
ok try the second so that you cn have confidence in yourself
anonymous
  • anonymous
do i need to make dy/dx a subject of the formula
anonymous
  • anonymous
try to arrange them first, start with dy/dx =
amistre64
  • amistre64
you can; y' just derives to y''; just be sure to use it as such in its product rule
anonymous
  • anonymous
yes thats the other way doing it,, either one will arrive at the right solution
amistre64
  • amistre64
to implicit it again would be easier; no quotient ruls to fight with i believe
anonymous
  • anonymous
ok thats rigth....try implicitly ..
amistre64
  • amistre64
5(x-5)^4 + 5(y-2)^4 y' =0 20(x-5)^3 + 5(y-2)^4 y'' + 20(y-2)^3 y' y' = 0 then if it works out that way :)
anonymous
  • anonymous
Dx(5(x-5)^4) + Dx(5(y-2)^4 dy/dx) =Dx(0).......ok amistre did it now hehehehe
amistre64
  • amistre64
i got no idea if its right :)... other than the obvious typos that is -20[(x-4)^3 + (y-2)^3 y'^2] y'' = ------------------------- right? 5(y-2)^3
amistre64
  • amistre64
-20/5 reduces..... and y' = whatever it equals lol
anonymous
  • anonymous
the answer is suppose to be: y''=(-256(X-5)^3)/(y-2)^9
anonymous
  • anonymous
dy/dx=-(x-5)^{4}/(y-2)^{4}\[d ^{2}y/dx ^{2}=-256(x-5)^{3}/(y-2)^{5}\]
amistre64
  • amistre64
it prolly simplifies to that then lol...
anonymous
  • anonymous
yeah ,... thats the answer to it Snenhlala....try to practice till you arrive to that answer
anonymous
  • anonymous
sorry the denominator for the second derivative should read (y-2)^9
anonymous
  • anonymous
did you used product rule or quotient rule to find 2nd derivative
anonymous
  • anonymous
i did mine in quotient rule
anonymous
  • anonymous
use the quotient rule and sub for dy/dx
anonymous
  • anonymous
mulitply each term by (y-2)^4 then factor -4(x-5)^3(y-2)^3 this will leave (x-5)^5+(y-2)^4 which equals 64 (from the original equation)
anonymous
  • anonymous
should read (x-5)^5+ (y-2)^5 which equals 64
anonymous
  • anonymous
should i start to multiply from the 1st derivative or from the origin equation?
anonymous
  • anonymous
multiply the second derivative
anonymous
  • anonymous
by (y-2)^4
anonymous
  • anonymous
y'= -(x-5)^4 / (y-2)^4 -4(y-2)^4 (x-5)^3 - 4(x-5)^4 (y-2)^3 y' y"= ----------------------------------- now subst. y ' in this equqtion ((y-2)^4)^2 -4(y-2)^4 (x-5)^3 - 4(x-5)^4 (y-2)^3 [-(x5^4)/(y-2)^4] y"= ------------------------------------------------------------- ((y-2)^4)^2 -4(y-2)^4 (x-5)^3 + 4(x-5)^8 (y-2) y"= ----------------------------------- (y-2)^8
anonymous
  • anonymous
-4(y-2)^4 (x-5)^3 + 4(x-5)^8/(y-2) y"= ----------------------------------- ... sorry its divided by y-2 (y-2)^8
anonymous
  • anonymous
-4(y-2)^5 (x-5)^3 + 4(x-5)^8 y"= ----------------------------------- (y-2)^9
anonymous
  • anonymous
-4 (x-5)^3 [(y-2)^5 + (x-5)^5] y"= ----------------------------------- (y-2)^9
anonymous
  • anonymous
-4 (x-5)^3 [64] y"= -------------- (y-2)^9 57 seconds ago
anonymous
  • anonymous
-256 (x-5)^3 y"= -------------- (y-2)^9
anonymous
  • anonymous
well i think thats it hehehe
anonymous
  • anonymous
thank you
anonymous
  • anonymous
w.c.
anonymous
  • anonymous
did you arrived in the same procedure?
anonymous
  • anonymous
and of course the same answer? lol
anonymous
  • anonymous
yes i did arrive but it too tricky
anonymous
  • anonymous
yeah kind of hehehe,,

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