SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS


SOLUTION 1 : Begin with x3 + y3 = 4 . Differentiate both sides of the equation, getting

D ( x3 + y3 ) = D ( 4 ) ,

D ( x3 ) + D ( y3 ) = D ( 4 ) ,

(Remember to use the chain rule on D ( y3 ) .)

3x2 + 3y2 y' = 0 ,

so that (Now solve for y' .)

3y2 y' = - 3x2 ,

and

$ y' = \displaystyle{ - 3x^2 \over 3y^2 } = \displaystyle{ - x^2 \over y^2 } $ .

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SOLUTION 2 : Begin with (x-y)2 = x + y - 1 . Differentiate both sides of the equation, getting

D (x-y)2 = D ( x + y - 1 ) ,

D (x-y)2 = D ( x ) + D ( y ) - D ( 1 ) ,

(Remember to use the chain rule on D (x-y)2 .)

$ 2 (x-y) \ D (x-y) = 1 + y' - 0 $ ,

2 (x-y) (1- y') = 1 + y' ,

so that (Now solve for y' .)

2 (x-y) - 2 (x-y) y' = 1 + y' ,

- 2 (x-y) y' - y' = 1 - 2 (x-y) ,

(Factor out y' .)

y' [ - 2 (x-y) - 1 ] = 1 - 2 (x-y) ,

and

$ y' = \displaystyle{ 1 - 2 (x-y) \over - 2 (x-y) - 1 } = \displaystyle{ 2y - 2x + 1 \over 2y - 2x - 1 } $ .

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SOLUTION 3 : Begin with $ y = \sin(3x + 4y) $ . Differentiate both sides of the equation, getting

$ D(y) = D ( \sin(3x + 4y) ) $ ,

(Remember to use the chain rule on $ D ( \sin(3x + 4y) ) $ .)

$ y' = \cos(3x + 4y) \ D ( 3x + 4y ) $ ,

$ y' = \cos(3x + 4y) ( 3 + 4 y' ) $ ,

so that (Now solve for y' .)

$ y' = 3 \cos(3x + 4y) + 4 y' \cos(3x + 4y) $ ,

$ y' - 4 y' \cos(3x + 4y) = 3 \cos(3x + 4y) $ ,

(Factor out y' .)

$ y' [ 1- 4 \cos(3x + 4y) ] = 3 \cos(3x + 4y) $ ,

and

$ y' = \displaystyle{ 3 \cos(3x + 4y) \over 1- 4 \cos(3x + 4y) } $ .

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SOLUTION 4 : Begin with y = x2 y3 + x3 y2 . Differentiate both sides of the equation, getting

D(y) = D ( x2 y3 + x3 y2 ) ,

D(y) = D ( x2 y3 ) + D ( x3 y2 ) ,

(Use the product rule twice.)

$ y' = \{ x^2 D ( y^3 ) + D ( x^2 ) y^3 \} + \{ x^3 D ( y^2 ) + D ( x^3 ) y^2 \} $ ,

(Remember to use the chain rule on D ( y3 ) and D ( y2 ) .)

$ y' = \{ x^2 ( 3y^2 y' ) + ( 2x ) y^3 \} + \{ x^3 ( 2 y y' ) + ( 3x^2 ) y^2 \} $ ,

y' = 3x2 y2 y' + 2x y3 + 2x3 y y' + 3x2 y2 ,

so that (Now solve for y' .)

y' - 3x2 y2 y' - 2x3 y y' = 2x y3 + 3x2 y2 ,

(Factor out y' .)

y' [ 1 - 3x2 y2 - 2x3 y ] = 2x y3 + 3x2 y2 ,

and

$ y' = \displaystyle{ 2x y^3 + 3x^2 y^2 \over 1 - 3x^2 y^2 - 2x^3 y } $ .

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SOLUTION 5 : Begin with $ e^{xy} = e^{4x} - e^{5y} $ . Differentiate both sides of the equation, getting

$ D(e^{xy} ) = D ( e^{4x} - e^{5y} ) $ ,

$ D( e^{xy} ) = D ( e^{4x} ) - D ( e^{5y} ) $ ,

$ e^{xy} D( xy ) = e^{4x} D ( 4x ) - e^{5y} D( 5y ) $ ,

$ e^{xy} ( xy' + (1) y ) = e^{4x} ( 4 ) - e^{5y} ( 5y' ) $ ,

so that (Now solve for $ y' $ .)

$ xe^{xy} y' + y e^{xy} = 4 e^{4x} - 5e^{5y} y' $ ,

$ xe^{xy} y' + 5e^{5y} y' = 4 e^{4x} - y e^{xy} $ ,

(Factor out $ y' $ .)

$ y' [ xe^{xy} + 5e^{5y} ] = 4 e^{4x} - y e^{xy} $ ,

and

$ y' = \displaystyle{ 4 e^{4x} - y e^{xy} \over xe^{xy} + 5e^{5y} } $ .

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SOLUTION 6 : Begin with $ \cos^2 x + \cos^2 y = \cos( 2x + 2y ) $ . Differentiate both sides of the equation, getting

$ D( \cos^2 x + \cos^2 y ) = D ( \cos( 2x + 2y ) ) $ ,

$ D( \cos^2 x ) + D ( \cos^2 y ) = D ( \cos( 2x + 2y ) ) $ ,

$ (2 \cos x) D ( \cos x ) + (2 \cos y) D ( \cos y ) = - \sin( 2x + 2y ) D ( 2x + 2y ) $ ,

$ 2 \cos x ( - \sin x ) + 2 \cos y ( - \sin y ) ( y' ) = - \sin( 2x + 2y ) ( 2 + 2y' ) $ ,

so that (Now solve for y' .)

$ - 2 \cos x \sin x - 2 y' \cos y \sin y = - 2 \sin( 2x + 2y) - 2 y' \sin( 2x + 2y) $ ,

$ 2 y' \sin( 2x + 2y) - 2 y' \cos y \sin y = - 2 \sin( 2x + 2y) + 2 \cos x \sin x $ ,

(Factor out y' .)

$ y' [ 2 \sin( 2x + 2y) - 2 \cos y \sin y ] = 2 \cos x \sin x - 2 \sin( 2x + 2y) $ ,

$ y' = \displaystyle{ 2 \cos x \sin x - 2 \sin( 2x + 2y) \over 2 \sin( 2x + 2y) - 2 \cos y \sin y } $ ,

$ y' = \displaystyle{ 2 [ \cos x \sin x - \sin( 2x + 2y) ] \over 2 [ \sin( 2x + 2y) - \cos y \sin y ] } $ ,

and

$ y' = \displaystyle{ \cos x \sin x - \sin( 2x + 2y) \over \sin( 2x + 2y) - \cos y \sin y } $ .

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SOLUTION 7 : Begin with $ x = \sqrt{ x^2 + y^2 } $ . Differentiate both sides of the equation, getting

$ D( x ) = D ( \sqrt{ x^2 + y^2 } ) $ ,

1 = (1/2)( x2 + y2 )-1/2 D ( x2 + y2 ) ,

1 = (1/2)( x2 + y2 )-1/2 ( 2x + 2y y' ) ,

so that (Now solve for y' .)

$ 1 = \displaystyle{ (1/2) (2) ( x + y y' ) \over \sqrt{x^2 + y^2} } $ ,

$ 1 = \displaystyle{ x + y y' \over \sqrt{x^2 + y^2} } $ ,

$ \sqrt{x^2 + y^2} = x + y y' $ ,

$ \sqrt{x^2 + y^2} - x = y y' $ ,

and

$ y' = \displaystyle{ \sqrt{x^2 + y^2} - x \over y } $ .

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SOLUTION 8 : Begin with $ \displaystyle{ x - y^3 \over y + x^2 } = x + 2 $ . Clear the fraction by multiplying both sides of the equation by y + x2 , getting

$ \displaystyle{ x - y^3 \over y + x^2 } (y + x^2 ) = (x + 2) (y + x^2 ) $ ,

or

x - y3 = xy + 2y + x3 + 2x2 .

Now differentiate both sides of the equation, getting

D ( x - y3 ) = D ( xy + 2y + x3 + 2x2 ) ,

D ( x ) - D (y3 ) = D ( xy ) + D ( 2y ) + D ( x3 ) + D ( 2x2 ) ,

(Remember to use the chain rule on D (y3 ) .)

1 - 3 y2 y' = ( xy' + (1)y ) + 2 y' + 3x2 + 4x ,

so that (Now solve for y' .)

1 - y - 3x2 - 4x = 3 y2 y' + xy' + 2 y' ,

(Factor out y' .)

1 - y - 3x2 - 4x = (3y2 + x + 2) y' ,

and

$ y' = \displaystyle{ 1 - y - 3x^2 - 4x \over 3y^2 + x + 2 } $ .

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SOLUTION 9 : Begin with $ \displaystyle{ { y \over x^3 } + { x \over y^3 } } = x^2y^4 $ . Clear the fractions by multiplying both sides of the equation by x3 y3 , getting

$ \Big\{ \displaystyle{ { y \over x^3 } + { x \over y^3 } } \Big\} ( x^3 y^3 ) = x^2 y^4 ( x^3 y^3 ) $ ,

$ \displaystyle{ { y x^3 y^3 \over x^3 } + { x x^3 y^3 \over y^3 } } = x^2 x^3y^4 y^3 $ ,

y4 + x4 = x5 y7 .

Now differentiate both sides of the equation, getting

D ( y4 + x4 ) = D ( x5 y7 ) ,

D ( y4 ) + D ( x4 ) = x5 D (y7 ) + D ( x5 ) y7 ,

(Remember to use the chain rule on D (y4 ) and D (y7 ) .)

4 y3 y' + 4 x3 = x5 (7 y6 y' ) + ( 5 x4 ) y7 ,

so that (Now solve for y' .)

4 y3 y' - 7 x5 y6 y' = 5 x4 y7 - 4 x3 ,

(Factor out y' .)

y' [ 4 y3 - 7 x5 y6 ] = 5 x4 y7 - 4 x3 ,

and

$ y' = \displaystyle{ 5 x^4 y^7 - 4 x^3 \over 4 y^3 - 7 x^5 y^6 } $ .

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SOLUTION 10 : Begin with (x2+y2)3 = 8x2y2 . Now differentiate both sides of the equation, getting

D (x2+y2)3 = D ( 8x2y2 ) ,

3 (x2+y2)2 D (x2+y2) = 8x2 D (y2 ) + D ( 8x2 ) y2 ,

(Remember to use the chain rule on D (y2 ) .)

3 (x2+y2)2 ( 2x + 2 y y' ) = 8x2 (2 y y' ) + ( 16 x ) y2 ,

so that (Now solve for y' .)

6x (x2+y2)2 + 6 y (x2+y2)2 y' = 16 x2 y y' + 16 x y2 ,

6 y (x2+y2)2 y' - 16 x2 y y' = 16 x y2 - 6x (x2+y2)2 ,

(Factor out y' .)

y' [ 6 y (x2+y2)2 - 16 x2 y ] = 16 x y2 - 6x (x2+y2)2 ,

and

$ y' = \displaystyle{ 16 x y^2 - 6x (x^2+y^2)^2 \over 6 y (x^2+y^2)^2 - 16 x^2 y } $ .

Thus, the slope of the line tangent to the graph at the point (-1, 1) is

$ m = y' = \displaystyle{ 16(-1) (1)^2 - 6(-1) ((-1)^2+(1)^2)^2 \over 6 (1) ((-1)^2+(1)^2)^2 - 16 (-1)^2 (1) } = \displaystyle{ 8 \over 8 } = 1 $ ,

and the equation of the tangent line is

y - ( 1 ) = (1) ( x - ( -1 ) )

or

y = x + 2 .

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SOLUTION 11 : Begin with x2 + (y-x)3 = 9 . If x=1 , then

(1)2 + ( y-1 )3 = 9

so that

( y-1 )3 = 8 ,

y-1 = 2 ,

y = 3 ,

and the tangent line passes through the point (1, 3) . Now differentiate both sides of the original equation, getting

D ( x2 + (y-x)3 ) = D ( 9 ) ,

D ( x2 ) + D (y-x)3 = D ( 9 ) ,

2x + 3 (y-x)2 D (y-x) = 0 ,

2x + 3 (y-x)2 (y'-1) = 0 ,

so that (Now solve for y' .)

2x + 3 (y-x)2 y'- 3 (y-x)2 = 0 ,

3 (y-x)2 y' = 3 (y-x)2 - 2x ,

and

$ y' = \displaystyle{ 3 (y-x)^2 - 2x \over 3 (y-x)^2 } $ .

Thus, the slope of the line tangent to the graph at (1, 3) is

$ m = y' = \displaystyle{ 3 (3-1)^2 - 2(1) \over 3 (3-1)^2 }
= \displaystyle{ 10 \over 12 } = \displaystyle{ 5 \over 6 } $ ,

and the equation of the tangent line is

y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,

or

y = (7/6) x + (13/6) .

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SOLUTION 12 : Begin with x2y + y4 = 4 + 2x . Now differentiate both sides of the original equation, getting

D ( x2 y + y4 ) = D ( 4 + 2x ) ,

D ( x2 y ) + D (y4 ) = D ( 4 ) + D ( 2x ) ,

( x2 y' + (2x) y ) + 4 y3 y' = 0 + 2 ,

so that (Now solve for y' .)

x2 y' + 4 y3 y' = 2 - 2x y ,

(Factor out y' .)

y' [ x2 + 4 y3 ] = 2 - 2x y ,

and

(Equation 1)

$ y' = \displaystyle{ 2 - 2x y \over x^2 + 4 y^3 } $ .

Thus, the slope of the graph (the slope of the line tangent to the graph) at (-1, 1) is

$ y' = \displaystyle{ 2 - 2(-1)(1) \over (-1)^2 + 4 (1)^3 }
= \displaystyle{ 4 \over 5 } $ .

Since y'= 4/5 , the slope of the graph is 4/5 and the graph is increasing at the point (-1, 1) . Now determine the concavity of the graph at (-1, 1) . Differentiate Equation 1, getting

$ y'' = \displaystyle{ (x^2 + 4 y^3) D(2 - 2xy) - (2 - 2xy) D(x^2 + 4 y^3) \over (x^2 + 4 y^3)^2 } $

$ = \displaystyle{ (x^2 + 4 y^3)((- 2x)y'+(- 2)y) - (2 - 2xy)(2x+12 y^2 y') \over (x^2 + 4 y^3)^2 } $ .

Now let x=-1 , y=1 , and y'=4/5 so that the second derivative is

$ y'' = \displaystyle{ [(-1)^2 + 4 (1)^3][(- 2(-1))(4/5)+(- 2)(1)] - [2 - 2(-1)(1)][2(-1)+12 (1)^2(4/5)]
\over ((-1)^2 + 4(1)^3)^2 } $

$ = \displaystyle{ (5) (8/5 - 2) - (4)(-2 + 48/5) \over 25 } $

$ = \displaystyle{ -2 - (152/5) \over 25 } $

$ = \displaystyle{ -162 \over 125 } $ .

Since y'' < 0 , the graph is concave down at the point (-1, 1) .

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Duane Kouba
1998-06-23