CPhill

I'am going to assume that this is :

y^3+sinh(x)y^2=3/2         ......if so.....we have

3y^2y ' + cosh(x)y^2  + 2sinh(x)yy '  =  0

y ' [ 3y^2  + 2y sinh(x) ]   =   - cosh(x)y^2

y'  = -cosh(x)y^2 / ' [ 3y^2  + 2y sinh(x) ]  =  - y*cosh(x)  / [3y + 2sinh(x)]

With this level of math question......I'm surprised at your lack of parentheses/brackets !!!

OK.....I see it, now.....

Not quite, Melody.......when you took the derivative of 3y^3,   it should be  9y^2 y'   instead of 6y^2 y'

Thanks, Melody.....I'll make the change....!!!

3y^3-4x^2y+xy=-5 

9y^2y'   - 8xy - 4x^2y' + y  + xy '   =  0 

y ' [ 9y^2 - 4x^2 + x] =  8xy + y

y ' =  y [ 8x  - 1]  /  [ 9y^2 - 4x^2 + x ]

Edit :  I copied it incorrectly, the first time....!!!!!

Let x be the number of quarters and y the number of dimes.....so...

x + y = 8   →   y = 8 - x     (1)

And the number of each type of coin times its value gives the total value

.25x + .10y =  1.25     (2)         put (1)  into (2)  for y

.25x + .10 [ 8 - x]  = 1.25         simplify

.25x + .80 - .10x  =  1.25

.15x + .80  = 1.25       subtract  .80  from each side

.15x  = .45      divide both sides by  .15

x = 3    this is the number of quarters

And using (1)     y =  8 - x   = 8 - 3  =   5   and this is the number of dimes

Melody......check out that website that Solveit referenced.....it includes step-by-step solutions  [ you must pay for this in WA ]

I just tried it.......pretty neat.....I'll definitely be adding that to my bookmarks....!!!

Note, Solveit

[ [9^2]^1 ]^0  = [ [81]^1 ] ^0   =  [81]^0  =  1

So

 [ [9^8]^7]^6] .........]^0    = [ 9^(8!)]^0   =  1

Thanks, Solveit for that website reference......I like the "step-by-step solution"  feature......!!!!

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 [9]                 =        9⁰        =     1

So......the first thing is infinitely larger  !!!!