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Find dy/dx by implicit differentiation. x^2-5xy+3y^2=7.
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High School Teacher
To find dy/dx by implicit differentiation. x^2-5xy+3y^2=7.
We differentiate the function f(x,y) = C both sides with respect to x and solve for dy/dx in terms of x and y, as y is not explicit.
x^2-5xy +3y^2 = 7.
We differentiate both sides with respect to x:
(x^2-5xy +3y^2)' = (7)'.
2x-5(1y+xdy/dx) +3*2y*dy/dx = 0.
2x-5y -5xdy/dx +6ydy/dx= 0.
We collect dy/dx terms together:
2x-5y +(-5x+6y)dy/dx = 0.
(6y-5x) dy/dx = 5y-2x.
we divide both sides by 6y-5x.
dy/dx = (5y-2x)/(6y-5x).
Therefore dy/dx = (5y-2x)/(6y-5x).
Posted by neela on December 6, 2010 at 2:03 AM (Answer #1)
Best answer as selected by question asker.
We'll differentiate the given expression both sides with respect to x:
(d/dx)( x^2-5xy+3y^2 = (d/dx)7
We'll differentiate each term from both sides:
(d/dx)( x^2) - (d/dx)5xy + (dy/dx)3y^2 = 0
2x - (5x*(dy/dx) + 5y) + 6y*(dy/dx) = 0
We'll factorize by (dy/dx):
(dy/dx)(5x + 6y) = -2x + 5y
We'll divide by 5x + 6y both sides:
dy/dx = (5y - 2x)/(5x + 6y)
Posted by giorgiana1976 on December 6, 2010 at 1:39 AM (Answer #2)
We are given the equation x^2-5xy+3y^2=7 and we have to find dy/dx.
Differentiating each term of x^2 - 5xy + 3y^2 = 7 with respect to x, we get
2x - 5x(dy/dx)- 5y + 6y(dy/dx) = 0
taking the terms with dy/dx to one side and the other terms to the opposite side we get
=> 2x - 5y = (dy/dx)( 5x - 6y)
divide both the sides by 5x - 6y
=> (dy/dx) = (2x - 5y) / ( 5x - 6y)
Therefore (dy/dx) = (2x - 5y) / ( 5x - 6y)
Posted by justaguide on December 6, 2010 at 1:40 AM (Answer #3)
High School Teacher
x^2 - 5xy + 3y^2 = 7
We need to differentiate the equation with respect to x.
==> dy/dx( x^2 - 5xy + 3y^2 ) = dy/dx ( 7)
==> dy/dx (x^2) - 5 dy/dx ( xy) + dy/dx ( 3y^2) = dy/dx ( 7).
We will differentiate with respect to x.
==> 2x - 5 ( 1*y + xy' ) + 6yy' = 0
==> 2x - 5y - 5xy'+ 6yy' = 0
Now we will combine the terms with y' on the left side.
==> 6yy' - 5xy' = 5y - 2x
Now we will factor y'.
==> y'*( 6y - 5x ) = 5y-2x
Now we will divide by (6y-5x).
==> y' = (5y-2x)/(6y-5x)
Posted by hala718 on December 6, 2010 at 1:42 AM (Answer #4)
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