how to solve using limit definition of the derivative for f(x)=2x^3+5x^2-7x-4
my ans:
6x^2+10x-7
plse correct me...
No - I thought you had gone through the rigmarole of using the limit definition and gotten that answer.
$\displaystyle f(x) = 2x^3 + 5x^2 -7x - 4$
$\displaystyle f(x + h) = 2(x + h)^3 + 5(x + h)^2 - 7(x + h) - 4$
$\displaystyle = 2(x^3 + 3x^2h + 3xh^2 + h^3) + 5(x^2 + 2xh + h^2) - 7x -7h -4$
$\displaystyle = 2x^3 + 6x^2h + 6xh^2 + 2h^3 + 5x^2 + 10xh + 5h^2 - 7x - 7h - 4$
So $\displaystyle f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 + 10xh + 5h^2 - 7h$
$\displaystyle = h[6x^2 + 6xh + 2h^2 + 10x + 5h - 7]$.
$\displaystyle \frac{f(x+h) - f(x)}{h} = 6x^2 + 6xh + 2h^2 + 10x + 5h - 7$
$\displaystyle \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = 6x^2 + 6x(0) + 2(0)^2 + 10x + 5(0) - 7$
$\displaystyle = 6x^2 + 10x - 7$.
Well you have to show your work probably and you do so by going through the grunt work of expanding and simplifying in order to evaluate the limit:
$\displaystyle \begin{aligned}f'(x) & = \lim_{h \to 0} \frac{2(x+h)^3+5(x+h)^2-7(x+h) - 4 \ - \ \left(2x^3 + 5x^2 - 7x - 4\right)}{h} \\ & = \lim_{h \to 0} \frac{2(x^3 + 2x^2h + 2xh^2 + h^3) +5(x^2 + 2xh + h^2) - \cdots}{h} \end{aligned}$
Eventually you can factor out an h from the top to cancel it with the bottom and you should get your answer. Like I said, a lot of grunt work.