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I am suppose to use the definition of a derivative to find $f'(2)$, where $f(x)= x^3 - 2x$

What do I do? I am not sure what they are asking, I can use the power rule and such but I know that is not what I am suppose to do.

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Do you know $f'(x)$ for any $x$ ? Then take $x=2$. If you don't, read the answer of Peter Tamarnoff. –  Student May 5 '12 at 0:52
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1 Answer 1

up vote 3 down vote accepted

We have $f(x) = x^3-2x$.

We build the quotient:

$$\frac{{f(x + h) - f\left( x \right)}}{h} = \frac{{{{\left( {x + h} \right)}^3} - 2\left( {x + h} \right) - \left( {{x^3} - 2x} \right)}}{h}$$

We have to expand the expression in the denominator. We get

$$\frac{{f(x + h) - f\left( x \right)}}{h} = \frac{{{x^3} + 3h{x^2} + 3{h^2}x + {h^3} - 2x - 2h - {x^3} + 2x}}{h}$$

We now cancel terms:

$$\frac{{f(x + h) - f\left( x \right)}}{h} = \frac{{3h{x^2} + 3{h^2}x + {h^3} - 2h}}{h}$$

Now, we distribute (divide) the $h$ in the denominator to all terms

$$\frac{{f(x + h) - f\left( x \right)}}{h} = \frac{{3h{x^2} + 3{h^2}x + {h^3} - 2h}}{h} = 3{x^2} + 3hx + h - 2$$

Now we can take the limit with no trouble:

$$\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} 3{x^2} + 3hx + h - 2 = 3{x^2} - 2$$

So $f'(x) = 3x^2-2$. So $f'(2) = 3\cdot 2^2-2 = 10$

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