Find the approximate change in $y$ as $x$ increases from 2 to 2.02 
Find the approximate change in $y$ as $x$ increases from 2 to 2.02 
  The equation of a curve is $y=4x^3-8x^2+10$
  a)Find $\frac{dy}{dx}$
  $\frac{dy}{dx}=12x^2-16x$
  But I don't know how to answer below  
  b) "Find the approximate change in $y$ as $x$ increases from 2 to 2.02" 
  I have try 
  $12(2)^2-16(2) = 16/2.02=$ not right 
  $4(2)^3-8(2)^2+10=10/2.02=$ not right
  $12(2.02)^2-16(2.02)=16.644/2=$ not right
  etc....
  the right answer is =$0.32$

help out thanks.
 A: We have $y=4x^3-8x^2+10$, and therefore $\frac{dy}{dx}=12x^2-16x$.
We use the tangent line approximation, also known as the linear approximation.
The derivative at $x=2$ is equal to $16$. Therefore, if $\Delta x$ represents the change in $x$, and $\Delta y$ represents the change in $y$, we have
$$\Delta y \approx (16)\Delta x.$$
Remarks: One important way to get insight about the linear approximation is geometric. Let $f(x)=4x^3-8x^2+10$. The idea is that the tangent line at $x=2$ is close to the curve when $x$ is close to $2$, that the tangent line kisses the curve at $x=2$.  A tiny bug, sitting on the curve $y=f(x)$ at $x=2$, would think she was sitting on a straight ine, the tangent line.
Recall that the tangent line at $x=a$ has equation
$$y-f(a)=f'(a)(x-a).$$
In our case, the tangent line has equation
$$y-f(2)=16(x-2).$$
Because the tangent line is close to the curve when $x$ is close to $2$, we have
$$f(2.02)-f(2)\approx (16)(2.02-2).$$
This says that the change in $y$ is approximately $(16)(0.02).$
Another way of thinking about it is kinematic, in terms of motion.  So let us use the letter $t$ instead of $x$. A particle is moving along the $y$-axis. At any time $t$, the displacement of the particle is $4t^3-8t^2+10$. Then the velocity at time $t$ is the derivative of $4t^3-8t^2+10$, evaluated at $t=2$.  If time changes from $2$ to $2.02$, then the change in $y$ (the change in displacement) is approximately the velocity at time $2$ times the elapsed time. So the change in $y$ is approximately $(16)(0.02)$.  The reason that the approximation is reasonable is that the velocity does not change very much from time $2$ to time $2.02$, so the velocity remains close to $16$. If the velocity were exactly $16$, then the change in displacement would be exactly $(16)(0.02)$. Since velocity does change a little, the approximation is not exact.
It is worthwhile to do an explicit numerical calculation to check  how good the tangent line approximation is in this case. The calculator says that $f(2.02)$ is nearly equal to $10.326432$, so to calculator accuracy, the change in $y$ is about $0.326432$.  The linear approximation we made predicts a change of approximately $0.32$. Pretty close!
Finally, we can think of our calculation in terms of the definition of the derivative. Recall that
$$f'(2)=\lim{h\to 0} \frac{f(2+h)-f(2)}{h}.$$
So if $h$ is kind of close to $0$, like $h=0.02$, then we should have
$$\frac{f(2+h)-f(2)}{h}\approx f'(2).$$
This can be written as $f(2+h)-f(2) \approx (f'(2))h$.
