1.11 Differentiability

$\newenvironment {prompt}{}{} \newcommand {\bart }[0]{BART} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We determine differentiability at a point

Differentiability

Differentiability. The function $f(x)$ is said to be differentiable at the point $x= x_0$ if the following limit exists: $\lim _{h\to 0} \frac {f(x_0+h) -f(x_0)}{h}.$ We denote the limit by $f'(x_0)$ , the derivative of $f$ at $x_0$ .

Geometrically, the derivative $f'(x_0)$ gives us the slope of a tangent line. Conceptually, it represents an instantaneous rate of change. In this section, we will explore the three main reasons that a function is not differentiable at a point. These are discontinuities, corner points, and vertical tangent lines.

If $f(x)$ is differentiable at $x = x_0$ , then $f(x)$ is continuous at $x=x_0$ , i.e., differentiability implies continuity.

An immediate consequence of this theorem is that if $f(x)$ is not continuous at $x = x_0$ , then it cannot be differentiable there.

If $f(x)$ is not continuous at $x = x_0$ then $f(x)$ is not differentiable at $x = x_0$ .

If the graph of a function has a removable, jump or infinite discontinuity, then the function is not differentiable at the corresponding point.

There are two other common reasons that a function might not be differentiable.

Corner Point The function $f(x)$ has a corner point at $x = x_0$ if the difference quotient has a jump discontinuity at $x = x_0$ , i.e., $\lim _{h\to 0^-} \frac {f(x_0 +h)-f(x_0)}{h} \text { and } \lim _{h\to 0^+} \frac {f(x_0 +h)-f(x_0)}{h}$ both exist, but are unequal.

The function $f(x) = |x|$ has a corner point at $x = 0$ . The left-hand limit, $\lim _{h\to 0^-} \frac {f(x_0 +h)-f(x_0)}{h} = \lim _{h\to 0^-} \frac {|h|}{h} = -1$ whereas the right-hand limit, $\lim _{h\to 0^+} \frac {f(x_0 +h)-f(x_0)}{h} = \lim _{h\to 0^+} \frac {|h|}{h} = 1.$

$\graph {abs(x)}$

The other common occurrence of a point of non-differentiability is at a vertical tangent line.

Vertical Tangent Line The function $f(x)$ has a vertical tangent line at $x = x_0$ if the vertical line $x = x_0$ is tangent to the graph of $y = f(x)$ at the point $(x_0, f(x_0))$ .

The function $f(x) = \sqrt [3] x$ has a vertical tangent line at $x = 0$ . $\graph {x^{1/3}}$

Below is the graph of $y = f(x)$ . Answer the questions below the graph. You can zoom in on the graph if you need to. $\graph {x^2}$ Is $f(x)$ differentiable at $x = 1$ ?

Yes No

If yes, is $f'(1)$ positive, negative or zero?

$f'(1) > 0$ $f'(1) < 0$ $f'(1) = 0$

Below is the graph of $y = f(x)$ . Answer the questions below the graph. You can zoom in on the graph if you need to. $\graph {3-abs(x-1)}$ Is $f(x)$ differentiable at $x = 1$ ?

Yes No

If no, why not?

discontinuity at $x = 1$ corner point at $x = 1$ vertical tangent line at $x = 1$

Below is the graph of $y = f(x)$ . Answer the questions below the graph. You can zoom in on the graph if you need to. $\graph {1 - (x-1)^{1/3}}$ Is $f(x)$ differentiable at $x = 1$ ?

Yes No

If no, why not?

discontinuity at $x = 1$ corner point at $x = 1$ vertical tangent line at $x = 1$

Below is the graph of $y = f(x)$ . Answer the questions below the graph. You can zoom in on the graph if you need to. $\graph {(x-2)^2}$ Is $f(x)$ differentiable at $x = 1$ ?

Yes No

If yes, is $f'(1)$ positive, negative or zero?

$f'(1) > 0$ $f'(1) < 0$ $f'(1) = 0$

Below is the graph of $y = f(x)$ . Answer the questions below the graph. You can zoom in on the graph if you need to. $\graph {1/(x-1)}$ Is $f(x)$ differentiable at $x = 1$ ?

Yes No

If no, why not?

discontinuity at $x = 1$ corner point at $x = 1$ vertical tangent line at $x = 1$

Below is the graph of $y = f(x)$ . Answer the questions below the graph. You can zoom in on the graph if you need to. $\graph {1 + (x-1)^2}$ Is $f(x)$ differentiable at $x = 1$ ?

Yes No

If yes, is $f'(1)$ positive, negative or zero?

$f'(1) > 0$ $f'(1) < 0$ $f'(1) = 0$

$\graph {2^x \left \{-2<x<2\right \}}$

$\graph {2^x \left \{-2<x<2\right \}, x \left \{2<x<4 \right \}}$

$\graph [color:red]{2^x}$

$\graph [style:dashed]{2^x}$