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I have the following calculus problem. How can I find the derivative of $f(x) = x^3 - 3x^2$ using the definition of the derivative?

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up vote 7 down vote accepted

Hint: Write out the difference quotient $f(x+h)-f(x)\over h$. Expand the numerator fully. Cancel the terms that you can in the numerator (terms involving $x^3$ and $3x^2\thinspace$). Once you've done this, every remaining term in the numerator will have a factor of $h$ in it. Factor out an $h$ from the numerator and cancel it with the $h$ in the denominator. This will leave you with only two terms in the numerator without a factor of $h$ and no $h$ in the denominator. Now you've set up the difference quotient so that finding the limit as $h$ tends to zero is easy. Of course, that limit is the derivative that you are looking for.

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Yeah been was along time since I did it. Thanks for that description – GivenPie Mar 4 '12 at 2:35

The definition is

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

I'll help you with the first one

$$\frac{{d\left\{ {{x^2}} \right\}}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2} - {x^2}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} + 2xh + {h^2} - {x^2}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + {h^2}}}{h} = \mathop {\lim }\limits_{h \to 0} \ (2x + h) = 2x$$

I guess with little effort you'll find $(x^3)'$ and maybe generalize it by induction to $(x^n)'$ using the chain rule.

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Thanks i got it – GivenPie Mar 4 '12 at 2:34

derivative of $x^n = nx^{n-1}$ apply that

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

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thanks, I used the formula and got the same answer – GivenPie Mar 4 '12 at 2:41

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