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

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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.**