Derivative interpretations [2.3]

Quiz on Friday, over sections 2.1-2.2.

(Change in $f$)= (change in $x$) $\times$ (rate of change)

...or "forecasting".

Using the derivative (change in a function per change in x) to estimate the function near a known value.

For a line:

The change in $x$ times the slope is the change in $y$: $$\begineq \Delta x \cdot m=&\Delta x \cdot \frac{\Delta y}{\Delta x}\\ =&\Delta y \endeq $$ And, $f_(x_2)=f(x_1)+\Delta y$ because $$\begineq f(x_1)+\Delta y =&f(x_1)+\left[ f(x_2)-f(x_1)\right]\\ =&f(x_2) \endeq $$ Putting these together: $$\begineq f(x_2)=&f(x_1)+\Delta y\\ =&f(x_1)+\Delta x\cdot m \endeq $$

The tangent-line approximation

In this diagram, what I called "$\Delta x$" above is now written as just "$\Delta$":

The line that is tangent to $f(x)$ at $x$ has slope $f'(x)$. As long as $f(x)$ is not changing too rapidly, then moving along the tangent line to $x+\Delta$ gets you to a point (in green) that has nearly the same $y$-coordinate as if you moved along $f(x)$ to the point (in black) at $x+\Delta$.

The tangent-line approximation: $$f(x+\Delta)\approx f(x)+\Delta\cdot f'(x)$$