y= x^5+ 2x^3+ 6. If dy/dx=0 , find x.

To calculate the roots of the equation dy/dx=0, we'll have
to differentiate the given relation both sides.

y= x^5+
2x^3+ 6

dy= (x^5+ 2x^3+
6)'dx

To differentiate the given expression x^5+ 2x^3+ 6,
we'll differentiate each term of this expression, with respect to
x.

(x^5+ 2x^3+ 6)' = (x^5)'+ (2x^3)'+
(6)'

To calculate the derivative of the power
function;

f(x) = x^n

f'(x) =
(x^n)'

But (x^n) = x*x*x*......*x, n
times

(x^n)' = (x*x*x*......*x)' = x'*x*...*x +
x*x'*x*...*x + ...+x*x*...*x' = n*x^(n-1)

If n = 5
=> (x^5)' = 5x^4

For n = 3 => (2x^3)' =
6x^2

(6)' = 0

(x^5+ 2x^3+ 6)'
= 5x^4 + 6x^2

Now, we'll calculate dy/dx = 0
<=> 5x^4 + 6x^2 = 0

We'll factorize by x^2
and we'll get:

x^2(5x^2 + 6)=
0

x^2 = 0

x1=x2 =
0

5x^2 + 6 = 0

We'll subtract
6 both sides:

5x^2 = -6
impossible!

x^2>0, for any value of
x

5x^2>0 also, so it's imposible for 5x^2 to have
negative values, for any x.

The
real solutions of dy/dx = 0 are x1=x2 = 0.