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