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

Let y = u/v such that:

u= 2x ==> u' = 2

v= (x^2 + 1) ==> v' = 2x

dy = (u'v-uv')/v^2

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

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

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

dy = 2(1-2x^2)/ (x^2 + 1)^2 = 0

==> (1-2x^2 = 0

==> 2x^2 = 1

==> x^2 = (1/2)

==> x1= sqrt(1/2) = sqrt2/2

==> x2= -sqrt(1/2) = -sqrt2/2

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

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

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

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

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

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

y' = 0 when 4 -x^2 = 0

4 = x^2 .

x = 2, or -2.

First, we'll have to differentiate the given function.

Because the function is a ratio, we'll calculate it's derivative using quotient rule:

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

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

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

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

We have, at numerator, a difference of squares:

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

(4-2x^2) = (2-x*sqrt2)(2+x*sqrt2)

Because the denominator of dy is always positive, for any value of x, only the numerator could be zero.

(2-x*sqrt2)(2+x*sqrt2)=0

We'll set each factor as zero.

2-x*sqrt2=0,

x*sqrt2=2

We'll divide by sqrt2:

x = 2/sqrt2

x = 2*sqrt2/2

**x1 = sqrt2**

2+x*sqrt2 = 0

x*sqrt2 = -2

We'll divide by sqrt2:

x = -2/sqrt2

**x2 = -sqrt2**

**The solutions of dy = 0 are : {-sqrt2 ; sqrt2}.**

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