using definition of derivative $f(x)=x^3-6x^2+9x-5$ is given.
What is the value of $$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h}$$ 
I tried to use the definition of derivative,and here it seems like the expression will be equal to something like the 2nd derivative of $f(x)$ but I'm confused with $2h$ and $-3h$.
 A: We have
$$f'(x)=3x^2-12x+9$$
and notice that
$$f'(1)=f'(3)=0$$
Now by the definition of the derivative we have
$$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h}=\lim_{u\to0}\frac{f'(1+u)-f'(1)}{u}-\frac{3}{2}\lim_{v\to0}\frac{f'(3+v)-f'(3)}{v}\\
=f''(1)-\frac{3}{2}f''(3)=-15 $$ 
A: Hints:


*

*$f'(1+ 2 h) = 9 - 12 (1 + 2 h) + 3 (1 + 2 h)^2$

*$f'(3-3h) =9-12 (3-3 h)+3 (3-3 h)^2$

*$f'(1+2h) + f'(3-3h) = 8-12 (3-3 h)+3 (3-3 h)^2-12 (1+2 h)+3 (1+2 h)^2$


So, we have:
$$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h} = \lim_{h\to0}\dfrac{-30 h + 39 h^2}{2h} = \lim_{h\to0}\dfrac{-30  + 39 h}{2} = -15$$ 
Can you fill in the details?
A: This is not the second derivative, since the points the first derivative is evaluated at are not the same points. That being said, you can just calculate the first derivative, plug in $1+2h, 3-3h$, and see what you get.
For reference, note that you'll get (after some simplification):
$$\lim_{h\to 0} \frac{3h(13h-10)}{2h}=\lim_{h\to 0} \frac{-30h}{2h}=-15$$
A: Hint: Note that $f'(1)=f'(3)=0$. Rewrite our limit as 
$$\lim_{h\to 0} \frac{f'(1+2h)-f'(1)}{2h}+\lim_{h\to 0}\left(-\frac{3}{2}\right)\frac{f'(3-3h)-f'(3)}{-3h}.$$
