Question

when you toast bread it losses its nutritional value ?​

Answer 1

Answer:

Toasting bread doesn't change its nutritional value, but it may decrease the glycemic index. Toasted bread calories aren't any fewer than untoasted bread calories. Toasting also doesn't affect carbohydrates or gluten; it may lower the glycemic index of bread, which is an advantage.

Answer 2

Begin with x2 + xy + y2 = 1 . Differentiate both sides of the equation, getting

D ( x2 + xy + y2 ) = D ( 1 ) ,

2x + ( xy' + (1)y ) + 2 y y' = 0 ,

so that (Now solve for y' .)

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

(Factor out y' .)

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

and the first derivative as a function of x and y is

(Equation 1)

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

To find y'' , differentiate both sides of this equation, getting

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

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

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

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

Use Equation 1 to substitute for y' , getting

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

(Get a common denominator in the numerator and simplify the expression.)

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

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

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

\displaystyle{ (x + 2y)^2 \over 1 } } $

$ = \displaystyle{ { -6x^2 - 3xy - 3xy - 6y^2 \over x+2y }

{ 1 \over (x + 2y)^2 } } $

$ = \displaystyle{ -6x^2 - 6y^2 - 6xy \over (x+2y)^3 } $ .

This answer can be simplified even further. Note that the original equation is

x2 + xy + y2 = 1 ,

so that

(Equation 2)

x2 + y2 = 1 - xy .

Use Equation 2 to substitute into the equation for y'' , getting

$ y'' = \displaystyle{ -6x^2 - 6y^2 -6xy \over (x+2y)^3 } $

$ = \displaystyle{ -6 (x^2 + y^2) - 6xy \over (x+2y)^3 } $

$ = \displaystyle{ -6 (1 - xy) - 6xy \over (x+2y)^3 } $

$ = \displaystyle{ -6 + 6xy - 6xy \over (x+2y)^3 } $ ,

and the second derivative as a function of x and y is

$ y'' = \displaystyle{ -6 \over (x+2y)^3 } $ .