Given that y=2x/(2+4x^2) determine the roots of dy.

Expert Answers
hala718 eNotes educator| Certified Educator

y= 2x/(2+4x^2)

First let us factor:

y= 2x/2(1+2x^2)

  = x/(1=2x^2)

Now leyt us differentiate y:

y= x/(1+2x^2) = u/v

u= x   ==> u' = 1

v= 1+2x^2 ==> v' = 4x

==> y'= (u'v- uv')/v^2

           = (1+2x^2 - 4x^2)/(1+2x^2)^2

             = (1-2x^2)/(1+2x^2)^2

Now dy = 0   when  (1-2x^2) = 0

==> 2x^2 = 1

==> x^2 = 1/2

==> x= +- sqrt(1/2) = +-sqrt2/2

==> the roots are:

x1= sqrt2/2    and   x2= -sqrt2/2

neela | Student

y = 2x/(2+4x^2)

To determine the roots of dy, we diffrentiate both sides and set dy = o and solve for x.

dy = {(2x)'(2+4x^2)- 2x(2+4x^2)'}/(2+4x^2).

dy = {2(2+4x^2)- 2x(2*4x)}/(2+4x^2)^2

dy = {4+8x^2 - 16x^2}/(2+4x^2)^2

dy = {4-8x^2}/(2+4x^2). Setting this to zero, we get:

4-8x^2 = 0

4 = 8x^2.

x^2 = 4/8 = 1/2

So x = sqrt(1/2) = sqrt2/2 or x= -sqrt(1/2) = -sqrt2/2 are the roots of the dy = 0

giorgiana1976 | Student

For the beginning, we'll have to differentiate the given function.

We notice that the function is a ratio and we'll calculate it's derivative using quotient rule:

(u/v)'= (u'*v-u*v')/v^2

dy = [2x/(4x^2+2)]'

dy = [(2x)'*(2+4x^2)-2x*(2+4x^2)']/(2+4x^2)^2

dy = (4 + 8x^2-16x^2)/(2+4x^2)^2

dy = (4-8x^2)/(2+4x^2)^2

We'll factorize by 4 both, numerator and denominator:

dy = 4(1-2x^2)/4(1+2x^2)^2

dy = (1-2x^2)/(1+2x^2)^2

We have, at numerator, a difference of squares:

a^2-b^2=(a-b)(a+b)

(1-2x^2) = (1-xsqrt2)(1+xsqrt2)

Because the denominator of f'(x) is always positive, for any value of x, only the numerator could be zero.

(1-xsqrt2)(1+xsqrt2) = 0

We'll set each factor as zero.

(1-xsqrt2) = 0

xsqrt2 = 1

x1 = sqrt2/2

(1+xsqrt2) = 0

xsqrt2 = -1

x2 = -sqrt2/2

The roots of dy = 0 are : {-sqrt2/2 ; sqrt2/2}.