Find the Explicit Solution S to Y 3cos

SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS

SOLUTION 1 : Begin with x ³ + y ³ = 4 . Differentiate both sides of the equation, getting

D ( x ³ + y ³ ) = D ( 4 ) ,

D ( x ³ ) + D ( y ³ ) = D ( 4 ) ,

(Remember to use the chain rule on D ( y ³ ) .)

3x ² + 3y ² y' = 0 ,

so that (Now solve for y' .)

3y ² y' = - 3x ² ,

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 = x ² y ³ + x ³ y ² . Differentiate both sides of the equation, getting

D(y) = D ( x ² y ³ + x ³ y ² ) ,

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

(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 ( y ³ ) and D ( y ² ) .)

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

y' = 3x ² y ² y' + 2x y ³ + 2x ³ y y' + 3x ² y ² ,

so that (Now solve for y' .)

y' - 3x ² y ² y' - 2x ³ y y' = 2x y ³ + 3x ² y ² ,

(Factor out y' .)

y' [ 1 - 3x ² y ² - 2x ³ y ] = 2x y ³ + 3x ² y ² ,

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)( x ² + y ² )^-1/2 D ( x ² + y ² ) ,

1 = (1/2)( x ² + y ² )^-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 + x ² , getting

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

x - y ³ = xy + 2y + x ³ + 2x ² .

Now differentiate both sides of the equation, getting

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

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

(Remember to use the chain rule on D (y ³ ) .)

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

so that (Now solve for y' .)

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

(Factor out y' .)

1 - y - 3x ² - 4x = (3y ² + 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 x ³ y ³ , getting

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

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

y ⁴ + x ⁴ = x ⁵ y ⁷ .

Now differentiate both sides of the equation, getting

D ( y ⁴ + x ⁴ ) = D ( x ⁵ y ⁷ ) ,

D ( y ⁴ ) + D ( x ⁴ ) = x ⁵ D (y ⁷ ) + D ( x ⁵ ) y ⁷ ,

(Remember to use the chain rule on D (y ⁴ ) and D (y ⁷ ) .)

4 y ³ y' + 4 x ³ = x ⁵ (7 y ⁶ y' ) + ( 5 x ⁴ ) y ⁷ ,

so that (Now solve for y' .)

4 y ³ y' - 7 x ⁵ y ⁶ y' = 5 x ⁴ y ⁷ - 4 x ³ ,

(Factor out y' .)

y' [ 4 y ³ - 7 x ⁵ y ⁶ ] = 5 x ⁴ y ⁷ - 4 x ³ ,

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 (x ²+y ²)³ = 8x ² y ² . Now differentiate both sides of the equation, getting

D (x ²+y ²)³ = D ( 8x ² y ² ) ,

3 (x ²+y ²)² D (x ²+y ²) = 8x ² D (y ² ) + D ( 8x ² ) y ² ,

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

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

so that (Now solve for y' .)

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

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

(Factor out y' .)

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

y = x + 2 .

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SOLUTION 11 : Begin with x ² + (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 ( x ² + (y-x)³ ) = D ( 9 ) ,

D ( x ² ) + 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 ) ) ,

y = (5/6) x + (13/6) .

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

D ( x ² y + y ⁴ ) = D ( 4 + 2x ) ,

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

( x ² y' + (2x) y ) + 4 y ³ y' = 0 + 2 ,

so that (Now solve for y' .)

x ² y' + 4 y ³ y' = 2 - 2x y ,

(Factor out y' .)

y' [ x ² + 4 y ³ ] = 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

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