# Math Insight

### Derivative from limit definition

Math 1241, Fall 2020
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Due date: Oct. 16, 2020, 11:59 p.m.
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Total points: 3
1. The limit definition of the derivative can be written as $f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} .$ We will use this definition to compute the derivative of the linear function $f(x)= 5 x + 7.$

Since $f(x+h) =$
and $f(x) =$
,
we can plug this results into the derivative definition:
$\displaystyle f'(x) = \lim_{h \to 0} [f(x+h)-f(x)]/h$
$\displaystyle = \lim_{h \to 0} \biggl[\biggl($
$\biggr) - \biggl($
$\biggr)\biggr]/h$

Next, we simplify the numerator and then perform the division by $h$:
$\displaystyle f'(x) =\lim_{h \to 0} \biggl[$
$\biggr]/h$
$\displaystyle =\lim_{h \to 0}$

We still have a limit in terms of $h$. However, the expression in the limit
depend on $h$. Therefore, the limit as $h$ goes to zero of this expression is
the expression itself. We can conclude that the derivative $f'(x)$ is
$f'(x)=$
.

2. Let $f(x)$ be the quadratic function $f(x)=5 x^{2}$.
1. Calculate the slope of the secant line from $x$ to $x+h$.

slope $= \biggl[\biggl($
$\biggr) - \biggl($
$\biggr)\biggr]\bigg/\biggl($
$\biggr)$

slope $= \biggl[$
$\biggr]\bigg/\biggl($
$\biggr)$

Cancel the $h$ in the numerator and denominator to get your final answer:
secant slope =

2. Calculate an expression for the derivative $\diff{f}{x}$ by taking the limit as $h \to 0$ of the secant slope calculated above. Since you've canceled the $h$ from the denominator, taking the limit is as simple as replacing $h$ with zero.

$\displaystyle \diff{f}{x} = \lim_{h \to 0 } \Bigl($
$\Bigr) =$

3. Sketch a graph of $f(x)$ and its derivative $\diff{f}{x}$. Pay attention to the places where $\diff{f}{x}$ is positive, zero, and negative.
Feedback from applet
correct function:
derivative matches function:

3. Let $f$ be the absolute value function $f(x)=|x|$. We'll explore where the function is differentiable (i.e., where the derivative exists). The function is differentiable at the point $a$ if the limit \begin{align*} \diff{f}{x}\biggr|_{x=a} = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{|a+h|-|a|}{h} \end{align*} exists. For the limit to exist, the expression must approach the same value for both positive and negative $h$.
1. For $a=-0.5$, calculate the slopes $\frac{f(a+h)-f(a)}{h}$ of the secant lines for:
$h=1$:
, $h= -1$:

$h=0.1$:
, $h= -0.1$:

$h=0.01$:
, $h= -0.01$:

Does the derivative $\diff{f}{x}\Bigr|_{x=-0.5}$ exist?
If so, what is its value?
Is the function differentiable at $x=-0.5$?

2. For $a=0$, calculate the slopes $\frac{f(a+h)-f(a)}{h}$ of the secant lines for:
$h=1$:
, $h= -1$:

$h=0.1$:
, $h= -0.1$:

$h=0.01$:
, $h= -0.01$:

Does the derivative $\diff{f}{x}\Bigr|_{x=0}$ exist?
Is the function differentiable at $x=0$?

3. Sketch a graph of the function $f(x)$ and its derivative $\diff{f}{x}$. Identify all points on the graph where the derivative fails to exist.
Feedback from applet
derivative:
function:
nondifferentiable points: