find d^7 y/dx^7 when y=7x^6+3x^4-5x^3+2
Answers
Step-by-step explanation:
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 } $ .
Click HERE to return to the list of problems.
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 } $ .
Click HERE to return to the list of problems.
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) } $ .
Click HERE to return to the list of problems.
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 } $ .
Click HERE to return to the list of problems.
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} } $ .
Click HERE to return to the list of problems.
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 } $ .
Click HERE to return to the list of problems.