Asked by Michelle
How do I differentiate
ln (2X^2 - 12X + Y^2 -10Y)
and the second differentiate it?
HELP ASAP is much appreciated
ln (2X^2 - 12X + Y^2 -10Y)
and the second differentiate it?
HELP ASAP is much appreciated
Answers
Answered by
Bosnian
Go on:
wolframalpha dot com
When page be open in rectangle type:
derivative ln (2X^2 - 12X + Y^2 -10Y)
and click option =
After a few seconds , when you see result click option:
Show steps
wolframalpha dot com
When page be open in rectangle type:
derivative ln (2X^2 - 12X + Y^2 -10Y)
and click option =
After a few seconds , when you see result click option:
Show steps
Answered by
Bosnian
second derivative is derivative of first derivative
on wolframalpha dot com type :
derivative 4 ( x - 3 ) / [ 2 x ^ 2 - 12 x + ( y -10 ) y ]
and click option =
then
Show steps
on wolframalpha dot com type :
derivative 4 ( x - 3 ) / [ 2 x ^ 2 - 12 x + ( y -10 ) y ]
and click option =
then
Show steps
Answered by
Reiny
If this were
2x^2 - 12x + y^2 - 10y = 0
it would be an ellipse and we could do something with it
I will assume that much
secondly I will assume you want dy/dx
4x - 12 + 2y dy/dx - 10dy/dx = 0
dy/dx (2y-10) = -4x + 12
dy/dx = (-4x + 12)/(2y-10) = (2x-6)/(5-y)
d(dy)/(dx)^2 = ( (5-y)(2) - (2x-6)(-dy/dx) )/(5-y)^2
I then substituted dy/dx with (2x-6)/(5-y) and got
( 2(5-y)^2 - (2x-6)^2 )/(5-y)^3
you could expand that but it would look messier.
2x^2 - 12x + y^2 - 10y = 0
it would be an ellipse and we could do something with it
I will assume that much
secondly I will assume you want dy/dx
4x - 12 + 2y dy/dx - 10dy/dx = 0
dy/dx (2y-10) = -4x + 12
dy/dx = (-4x + 12)/(2y-10) = (2x-6)/(5-y)
d(dy)/(dx)^2 = ( (5-y)(2) - (2x-6)(-dy/dx) )/(5-y)^2
I then substituted dy/dx with (2x-6)/(5-y) and got
( 2(5-y)^2 - (2x-6)^2 )/(5-y)^3
you could expand that but it would look messier.
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