DifferentiationWatch

Announcements
This discussion is closed.
Thread starter 14 years ago
#1
How would I differentiate:

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

The answer is: dy/dx = (3x^2 - y)/(2 + x)

Also:
squ.rt.(xy) + x + y^2 = 0

The answer is: dy/dx = -(2(xy)^(0.5) + y)/(4y(xy)^(0.5) + x)

Please show me how to do it! Thank You!
0
14 years ago
#2
cross multiply to get x^3 - xy = 2y --> 2y+xy=x^3 --> y(2+x)=x^3 --> y=x^3/(2+x)
Differentiate this as a quotient to get.. dy/dx= ((2+x).3x^2 - x^3)/(2+x)^2
Since x^3 = y(2+x), substitute for x^3 to get
dy/dx = ((2+x).3x^2 - (2+x)y)/(2+x)^2
(2+x) cancels out leaving
dy/dx = (3x^2 - y)/(2+x)
0
Thread starter 14 years ago
#3
Also:
squ.rt.(xy) + x + y^2 = 0

The answer is: dy/dx = -(2(xy)^(0.5) + y)/(4y(xy)^(0.5) + x)

Please show me how to do it! Thank You!
0
14 years ago
#4
see attached
edit: there's a typo on the first line.. it should be xy` + y. i hope you see it.
0
14 years ago
#5
Q). Differentiate wrt x: (xy)^(1/2) + x + y^2 = 0

---> x^(1/2).y^(1/2) + x + y^2 = 0
---> y^(1/2).(x^-1/2)/2 + x^(1/2).dy/dx.(y^-1/2)/2 + 1 + 2y.dy/dx = 0
---> y^(1/2)/[2x^(1/2)] + dy/dx[x^(1/2)]/[2y^(1/2)] + 1 + 2y.dy/dx = 0
---> dy/dx[x^(1/2)]/[2y^(1/2)] + 2y.dy/dx = -y^(1/2)/[2x^(1/2)] - 1
---> dy/dx{[x^(1/2)]/[2y^(1/2)] + 2y} = -{y^(1/2)/[2x^(1/2)] + 1}
---> dy/dx{[x^(1/2) + 4y^(3/2)]/[2y^(1/2)]} = -{y^(1/2)/[2x^(1/2)] + 1}
---> dy/dx = -{2y^(1/2)[y^(1/2)/2x^(1/2) + 1]}/[x^(1/2) + 4y^(3/2)]
---> dy/dx = -{y/x^(1/2) + 2y^(1/2)}/[x^(1/2) + 4y^(3/2)]
---> dy/dx = -{y + 2.x^(1/2).y^(1/2)}/[x + 4.x^(1/2).y^(3/2)]
---> dy/dx = -{2(xy)^(1/2) + y}/[x + 4y.x^(1/2).y^(1/2)]
---> dy/dx = -{2(xy)^(1/2) + y}/[x + 4y(xy)^(1/2)]

QED

Nima
0
X
new posts Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• University of Bradford
Thu, 24 Oct '19
• Cardiff University
Sat, 26 Oct '19
• Brunel University London
Sat, 26 Oct '19

Poll

Join the discussion

Yes (65)
23.3%
No (214)
76.7%