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

1 Answer
Jul 29, 2015

#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.