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.
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:
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.