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Find dy/dx by implicit differentiation. x^2-5xy+3y^2=7.

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drole | Student, Grade 10 | (Level 1) Honors

Posted December 6, 2010 at 1:39 AM via web

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Find dy/dx by implicit differentiation. x^2-5xy+3y^2=7.

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neela | High School Teacher | (Level 3) Valedictorian

Posted December 6, 2010 at 2:03 AM (Answer #1)

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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).

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 6, 2010 at 1:39 AM (Answer #2)

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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)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 6, 2010 at 1:40 AM (Answer #3)

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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)

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted December 6, 2010 at 1:42 AM (Answer #4)

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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)

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