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

Jul 29, 2015

$f ' \left(x\right) = 3 {x}^{2} - 6 x$

#### 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:

$f \left(x\right) = {x}^{n} \to f ' \left(x\right) = n {x}^{n - 1}$ where $n$ is any number.

For the function $f \left(x\right) = {x}^{3} - 3 {x}^{2} - 1$, 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 ($- {x}^{0} \to 0 \cdot - {x}^{-} 1 \to 0$). Now for ${x}^{3}$ and $- 3 {x}^{2}$, both involve the Power Rule's generalized form as mentioned earlier. So:

${x}^{3} \to 3 {x}^{3 - 1} \to 3 {x}^{2}$
$- 3 {x}^{2} \to - 3 \cdot 2 \cdot {x}^{2 - 1} \to - 6 x$

Thus by putting it together,
$f ' \left(x\right) = \frac{d \left(f \left(x\right)\right)}{\mathrm{dx}} = \frac{d \left({x}^{3} - 3 {x}^{2} - 1\right)}{\mathrm{dx}} = 3 {x}^{2} - 6 x$.

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

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

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.