help me palese....in calculus..implicit differentiation..tnx?

Question

1.srqareroot of x+y +xy=21
2.b^2x^2 + a^2y^2=a^2b^2

then wats the right answer

help..~!!!!!!!

Answer 1

You can put that in explicit form and then differenziate as usually.

Answer 2

1. Just differentiate each term with respect to x:

dx/dx + dy/dx + d(xy)/dx = 0 Righthand side zero because it's the derivative of a constant.
Use the product rule for
d(xy)/dx = (dx/dx)*y + x*(dy/dx)
Now I'm sure you know dx/dx = 1, and
dy/dx + x*(dy/dx) =(1 + x)(dy/dx)

Move to the righthand side all terms which don't contain dy/dx, and solve the equation for dy/dx.

2. Same idea. b and a are constants, so just treat them the same as if they were 5 or 8 etc.
Use the "chain rule" for
d(y^2)/dx = (d(y^2)/dy)*dy/dx
I don't think I need to tell you any more. If you still can't do it for yourself email [e-mail]

Answer 3

1. √(x+y +xy) = 21
(1 + dy/dx + y + xdy/dx)/[2√(x+y +xy)] = 0
So 1 + dy/dx + y + xdy/dx = 0
ie (1 + x)dy/dx = -(1 + y)
dy/dx = -(1 + y)/(1 + x)

2. b²x² + a²y² = a²b²
b² * 2x + a² * 2y * dy/dx = 0
2a²y * dy/dx = -2b²x
dy/dx = -(b²x)/(a²y)

Answer 4

Uh,Yeah you need learn how to spell first

Answer 5

1.1+y'+y+xy'=0
y'(1+x)=-(1+y)
y'=-(1+y)/(1+x)

2.2b^2x+2a^2yy'=0
2a^2yy'=-2b^2x
y'=-2b^2x/2a^2y
=-b^2x/a^2y