y = 4 x^3 + 6/3 x
Enter the derivative: y'=???
what's up? Simple power rule:
12x^2 + 6/3
the equation is
4x^3 + 6/3x (4x cubed plus 6 over 3x)
ah- a few parentheses would help
(4x^3 + 6)/3x
just divide by 3x:
4/3 x^2 + 6/(3x) = 4/3 x^2 + 2/x
f' = 8/3 x - 2/x^2
Or, using the quotient rule,
[(12x^2)(3x) - (4x^3+6)(3)]/(3x)^2
= (36x^3 - 12x^3 - 18) / 9x^2
= (24x^3-18) / 9x^2
= (8x^3-6) / 3x^2
= 8/3 x - 2/x^2
To find the derivative of the given function, we can use the power rule of derivatives. The power rule states that if we have a function of the form f(x) = x^n, then the derivative of f(x) with respect to x is given by f'(x) = n * x^(n-1).
In the given function, y = 4x^3 + (6/3)x. Let's take the derivative term by term:
For the first term, 4x^3, we use the power rule. The power rule tells us that the derivative of x^n is n * x^(n-1). In this case, n = 3, so the derivative of 4x^3 with respect to x is:
dy/dx = 4 * 3 * x^(3-1) = 12x^2.
For the second term, (6/3)x, we can simplify it to 2x. The derivative of 2x with respect to x is simply:
dy/dx = 2.
Therefore, the derivative of y = 4x^3 + (6/3)x is:
dy/dx = 12x^2 + 2.