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:
    $\displaystyle f'(x) = \lim_{h \to 0} [f(x+h)-f(x)]/h$
    $\displaystyle = \lim_{h \to 0} \biggl[\biggl($
    $\biggr) - \biggl($

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

  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 $= \biggl[\biggl($
      $\biggr) - \biggl($

      Simplify your expression by multiplying out the quadratic:
      slope $= \biggl[$

      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:
      , $h= -1$:

      , $h= -0.1$:

      , $h= -0.01$:

      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:
      , $h= -1$:

      , $h= -0.1$:

      , $h= -0.01$:

      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
      nondifferentiable points: