What is x if dy/dx=0 and y= 9x^4-3x^3-7?

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

y=
9x^4-3x^3-7

dy=
(9x^4-3x^3-7)'dx

To differentiate the given expression
9x^4-3x^3-7, we'll differentiate each term of this expression, with respect to
x.

(9x^4-3x^3-7)' = (9x^4)'- (3x^3)'-
(7)'

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 = 4
=> (9x^4)' = 4*9x^3

(9x^4)' =
36x^3

For n = 3 => (3x^3)' =
3*3x^2

(3x^3)' = 9x^2

(7)' =
0

(9x^4-3x^3-7)' = 36x^3 -
9x^2

Now, we'll calculate dy/dx = 0 <=> 36x^3
- 9x^2= 0

We'll factorize by 9x^2 and we'll
get:

9x^2(4x-1)= 0

We'll put
each factor as zero:

9x^2 =
0

x1 = 0

x2 =
0

4x-1 = 0

We'll add 1 both
sides:

4x = 1

x3 =
1/4

The real solutions of dy/dx = 0 are x1=
0, x2 = 0 and x3 = 1/4.