Question

How do you find the derivative of the function: `f(x) = x^3 - 3x^2 - 1`?

Answer 1

`f'(x)=3x^2-6x`

Explanation:

This function involves the most commonly used rule for taking derivatives: the Power Rule. It states that whenever you have a power for a variable, that is automatically considered a product of the variable with the power being subtracted by one. Here is a general form:

`f(x)=x^n->f'(x)=nx^(n-1)` where `n` is any number.

For the function `f(x)=x^3-3x^2-1`, the key is finding the variables in which you can take the derivative with respect to `x` (which the problem asks, but for others they should indicate which one if there are two variables).

Every time you have constants like `-1`, it automatically goes to zero when you take the derivative (`-x^0->0*-x^-1->0`). Now for `x^3` and `-3x^2`, both involve the Power Rule's generalized form as mentioned earlier. So:

`x^3->3x^(3-1)->3x^2`
`-3x^2->-3*2*x^(2-1)->-6x`

Thus by putting it together,
`f'(x)=(d(f(x)))/dx=(d(x^3-3x^2-1))/dx=3x^2-6x`.

If you use the Limit Definition with `f(x+h)` as the next point of the function and `f(x)` as the original for the instantaneous rate of change:

`f'(x)=lim_(h->0)(f(x+h)-f(x))/(h)`

Solving for the equation gives you the same answer (though a long work process). The Power Rule and other rules for derivatives give you the shortcuts to solve for even the most complicated problems in using the Limit Definition.