IMPLICIT DIFFERENTIATION

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Ryan Rigdon

Junior Member
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Jun 10, 2010
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FIND DxY BY IMPLICIT DIFFERENTIATION.

XY + COS(XY) = -4


MY WORK SO FAR.

D/DX (XY+COS(XY)) = D/DX (-4)

D/DX(XY) + D/DX(COS(XY)) = D/DX (-4)

D/DX (-4) = 0

D/DX (XY) = X D/DX(Y) + Y D/DX(X) = X DY/DX + Y

D/DX (COS(XY)) = ?

I AM STUCK HERE CAN SOME PLEASE HELP ME FINISH THIS PROBLEM.
 
Ryan Rigdon said:
FIND DxY BY IMPLICIT DIFFERENTIATION.

XY + COS(XY) = -4


MY WORK SO FAR.

D/DX (XY+COS(XY)) = D/DX (-4)

D/DX(XY) + D/DX(COS(XY)) = D/DX (-4)

D/DX (-4) = 0

D/DX (XY) = X D/DX(Y) + Y D/DX(X) = X DY/DX + Y

D/DX (COS(XY)) = (Using chain rule) ? = -Sin (XY) * D/DX (XY) .... continue....

I AM STUCK HERE CAN SOME PLEASE HELP ME FINISH THIS PROBLEM.
Click to expand...
 
sorry about the caps. thanx for the hint will report my findings in 15 minutes or so.
 
FIND DxY BY IMPLICIT DIFFERENTIATION.

XY + COS(XY) = -4


MY WORK SO FAR.

D/DX (XY+COS(XY)) = D/DX (-4)

D/DX(XY) + D/DX(COS(XY)) = D/DX (-4)

D/DX (-4) = 0

D/DX (XY) = X D/DX(Y) + Y D/DX(X) = X DY/DX + Y

D/DX (COS(XY)) = -sin(xy)*d/dx(xy) = -ysin(xy)

i then plugged everything back in and received the following answer.

xdy/dx + y - ysin(xy) = 0

x dy/dx = -y + ysin(xy)

dy/dx = (-y(1 - sin(xy)))/x in its simplified form.

does this answer the question of finding DxY by implicit differentiation?
 
Ryan Rigdon said:
FIND DxY BY IMPLICIT DIFFERENTIATION.

XY + COS(XY) = -4


MY WORK SO FAR.

D/DX (XY+COS(XY)) = D/DX (-4)

D/DX(XY) + D/DX(COS(XY)) = D/DX (-4)

D/DX (-4) = 0

D/DX (XY) = X D/DX(Y) + Y D/DX(X) = X DY/DX + Y

D/DX (COS(XY)) = -sin(xy)*d/dx(xy) = -ysin(xy) <<<< That's not correct = - sin(xy) * [x * dy/dx + y]

i then plugged everything back in and received the following answer.

xdy/dx + y - ysin(xy) = 0

x dy/dx = -y + ysin(xy)

dy/dx = (-y(1 - sin(xy)))/x in its simplified form.

does this answer the question of finding DxY by implicit differentiation?
Click to expand...
 
I was doing this whole problem wrong i think, i redid it and have the following answer which i think is much better tha my previous work

the problem once again

Find DxY by implicit differentiation

xy + cos(xy) = -4

my work

xDxY + y - sin(xy)*(xDxY + y) = 0

xDxY+ y -xsin(xy)DxY - ysin(xy) = 0

xDxY - xsin(xy)DxY = -y + ysin(xy)

DxY = (-y + ysin(xy))/(x - xsin(xy)) = (-y(1-sin(xy))/(x(1-sin(xy)) = -y/x

So by implicit differentiation xy + cos(xy) = -4, DxY = -y/x.


Correct?
 
Ryan Rigdon said:
I was doing this whole problem wrong i think, i redid it and have the following answer which i think is much better tha my previous work

the problem once again

Find DxY by implicit differentiation

xy + cos(xy) = -4

my work

xDxY + y - sin(xy)*(xDxY + y) = 0

xDxY+ y -xsin(xy)DxY - ysin(xy) = 0

xDxY - xsin(xy)DxY = -y + ysin(xy)

DxY = (-y + ysin(xy))/(x - xsin(xy)) = (-y(1-sin(xy))/(x(1-sin(xy)) = -y/x

So by implicit differentiation xy + cos(xy) = -4, DxY = -y/x.


Correct?
Click to expand...

Another way

Let

? = xy

then

d/d? [? + cos(?)] * d?/dx = 0

[1 - sin(?)] * [y + x*y'] = 0

when (1- sin?) <> 0, we have

y + xy' = 0

y' = -y/x
 
thanx Subhotosh Khan. its great that this site is here, as i have said before. peace bro.
 
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