Can someone help find the second derivative of this Calculus problem?

Question

The problem is:

y^2 + 2xy = 16

I want someone to show me the steps to get the FIRST and the SECOND Derivative, ... Please.

Thanks

Answer 1

respect to eiter one doesnt matter just pick one and do it (message to the first guy)

now:
we will do respect x since thats what we are use to the most

re write the equation as y^2 = -2xy, screw the 16 as its a constant- making it easier to look at

take the derviative of both sides: 2y (dy/dx) = -2(y +x(dy/dx))

cancel out 2's: y(dy/dx) = -y - x(dy/dx)

solve for dy/dx = -y/(y+x), this will come in hand later

take the derviative of this using qouteint rule
(dy^2/dx^2) = [(y+x)(dy/dx) - (-y)((dy/dx)+1)] / (y^2)

if anybody else repeats what i did on the below answers they just copied me

simplify subsituting (dy/dx) with what we found it to be (-y/(y+x))

-just simplify from here- but im pretty sure the teacher will take it from here as the next few steps are just simplifying

= -y/(x+y)^2 + y/(x+y)^2 - y^2/(x+y)^3
= -y^2/(x+y)^3

Answer 2

ok. on the 1st you utilize the quotient rule z' = ](a million+x^2)*0 -2*2x]/(a million+x^2)^2 = -4x/(a million+x^2)^2 z" follows the comparable technique. [(a million+x^2)^2*-4 - (-4x)*2*(a million+x^2)*2x]/(a million+x^2)^4. The numerator may well be factored to (a million+x^2)*[-4(a million+x^2) +16x]/(a million+x^2)^4. y' = 4*[(3x^2+4)^3]*6x = 24x*(3x^2+4)^3. Now use the product rule and chain rule to get y'' y" = 24x*[3*(3x^2+4)^2]*6x + 24*(3x^2+4)^3 = {24*(3x^2+4)^2}*[21x^2+4] it is tedious to assert the least.

Answer 3

Assuming a third independent variable t
i.e. y = y(t), x = x(t)

y^2 + 2 x y = 16

diff: w.r.t. t

2y y' + 2 x y' + 2y x' = 0

yy' + xy' + yx' =0

x' = x y'/y - y'
y' = - (y+x + yx')

Now solve by assigning y'=1 or x'=1, depending on
y = t or x = t

or either x' = x/y - 1
or y' = -(2y+x)

Coming to second derivative:
yy' + xy' + yx' =0

y'y' + yy'' + x'y' + xy'' + y'x' + yx'' =0

y' ^2 + yy'' + 2 x'y' + xy'' + yx'' = 0

Now assign y''=0 if y=t, or if x''=0 if x=t

if y''=0, y'^2 + 2 x'y' + x y'' = 0, and use x',y' from above
if x''=0, y'^2 + yy'' + 2x'y' +x y''=0, and use x',y' from above

Answer 4

Go to Math.com and enter the problem ands it will show you all the steps. It is a great site to learn and find anwsers tp problems and slove any type of equation. Let me know at [e-mail]. Then if not satisfied email me and I will help you. You can slove it !

Answer 5

Differentiate with respect to x,
2yy' + 2y + 2xy' = 0
y' = -y/(x+y)
y"
= -y'/(x+y) + y(1+y')/(x+y)^2
= -y/(x+y)^2 + y/(x+y)^2 - y^2/(x+y)^3
= -y^2/(x+y)^3
-----------
Evan,
Your work on y" is not right. Check it.

Answer 6

Derivative with respect to what? I presume you want to find dy/dx.

Differentiate implicitly.

2y(y') + (2y + 2xy') = 0

y'(2y + 2x) = -2y
y' = -y/(x+y)

y'' = -y'(x+y) - (-y)(1 + y') = -(-y/(x+y)) - -y(1 + -y/(x+y))

Anyway. Check my work.

Answer 7

2yy'+2xy'+2yx'=0
yy'+xy'+yx'=0 (first derivative)
you can simplify it like this
y'(x+y)= -yx'
y'= (-yx')/(x+y)
yy''+y'^2+x'y'+xy''+y'x'+yx''=0 (second derivative)
you can simplify it like this
y''(y+x)= -(y'^2+y'x'+yx'')
y''= -(y'^2+y'x'+yx'')/(y+x)

Answer 8

Differentiate w.r.t.x:-
2y.(dy/dx) + 2 y + (dy/dx).(2x) = 0
(2x + 2y).(dy/dx) = - 2y
dy/dx = - y / (x + y)
d²y / dx² :-
(x + y).(- dy/dx) + y.(1 + dy/dx) / [ x + y ]²
[-x.dy/dx + y ] / [ x + y]²
[ xy / (x + y) + y ] / [x + y ]²
[ (xy + y.(x + y) ) / (x + y)] / [ x + y ]²
[ 2xy + y² ] / [ x + y ]³

Answer 9

with respect to x or y?