Instantaneous Rate of Change/Derivative I am having a bit of trouble with a question on my Calc HW. Given P(x)= 3x^2+3, estimate the Instantaneous rate of change at x=5. This is what I have so far.
$$f(x+h) = 3(x+h)^2+3-(3x^2+3)$$
      $$= (3x+3h)(x+h)+3-(3x^2+3)$$
       $$= 3x^2+3xh+3xh+3h^2+3-(3x^2+3)$$
       $$= 3x^2+6xh+3h^2+3-3x^2-3$$
      $$ = (6xh+3h^2/h)$$
      $$ = h(6x+3h)/h$$
       eliminate $h$ and left with $6x+3h$
I know this is incorrect, but don't know where I went wrong. I know for the last step, you have to plug in 5 to x and solve to get the Rate of change.
 A: $$\frac{6xh+3h^2}{h}=6x+3h
$$  
Which approaches $6x$ as $h$ approaches $0$. 
Plug in $x=5$:  
$$6\times 5=30$$
Sorry to be so brief. I'm on a smartphone.
What you've got looks good, although it's difficult to read until it's formatted... I think that's in the works.  
Make your first line $$\frac {f(x+h) - f(h)}{h}$$
A: Your working is correct, you just need to take the limit $h\to 0$. Given the question asks you to 'estimate' the instantaneous rate of change, it sounds like they don't want you to find the derivative. Since you have attempted finding the derivative, the full solution would be:
$$
\begin{split} f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{x+h-x} \\ &=\lim_{h \to 0} \frac{3(x+h)^2+3-3x^2-3}{h} \\ &= \lim_{h\to 0}\frac{3(x^2+2xh+h^2)-3x^2}{h} \\ &= \lim_{h\to 0} \frac{3x^2+6xh+3h^2-3x^2}{h} \\ &=\lim_{h\to 0}\frac{6xh+h^2}{h} \\ &=\lim_{h\to 0}(6x+3h)=6x \end{split}
$$
So the instantaneous rate of change at $x=5$ is $f'(5)=6\times 5=30$. You can approximate this without the derivative by just choosing two points on the curve close to $5$ and finding gradient of the line between them. For example, choose the points $(5,78)$ and $(5.1, 81.03)$. The gradient of the line between them is given by:
$$
m=\frac{y_2-y_1}{x_2-x_1}=\frac{81.03-78}{5.1-5}=\frac{3.03}{0.1}=30.3
$$
which is approximately the instantaneous rate of change at $x=5$.
