# Differentiability

(Difference between revisions)
 Revision as of 09:46, 25 June 2012 (edit)← Previous diff Revision as of 12:49, 2 July 2013 (edit) (undo)m Next diff → Line 1: Line 1: + {{Image Description + |ImageName=A Differentiable Function + |Image=Math images 4.jpg + |ImageIntro=$y=x^2$ is an example of a function that is differentiable everywhere. + |ImageDescElem=If we were to draw a tangent through every point on the curve in the main image, we would not encounter any difficulty at any point because there are no discontinuities, sharp corners and straight vertical portions at any point. This means that the function is '''differentiable'''. + |ImageDesc=A function is differentiable at a point if it has a tangent at every point. That is, a function is differentiable at $x$ if the [[Limit|limit]] + + $\lim_{a \to 0} {{f(x+a)-f(x)} \over\ a}$ exists. + + It fails to be differentiable if: + + *$f(x)$ is not continuous at $a$ + + {{hide|[[Image:math_images_5.jpg|left|thumb| The limit does not approach the same value from the right as it does from the left, so the function is not differentiable]] + + + For example $f(x)= {1 \over\ x}$ is not continuous at $x=0$. The function is undefined at that point, hence it is not differentiable. + Computing the limit: + + + $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {{{1 \over\ a} -0} \over\ a}$ + $= \lim_{a \to 0} {1 \over\ a^2}$ + + + As $a$ approaches $0$ from the left and right,the denominator becomes smaller and smaller, hence the limit approaches $- \infty$ and $\infty$ respectively. The limits from the right and left are different so the limit does not exist hence the function is not differentiable.}} + + + *The graph has a sharp corner at $a$ + + {{hide|The function $f(x)=|x|$ has a sharp corner at $x=0$. [[Image:Absolute_value.png|right|thumb| http://commons.wikimedia.org/wiki/File:Absolute_value.png]] + + Computing the limit: + $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {{|a|-0} \over\ a} + = \lim_{a \to 0} {{|a|} \over\ a}$ + + As $a$ approaches $0$ from the right, the ratio is $1$, from the left, the ratio is $-1$. The limits are different, so the function is not differentiable.}} + + *The graph has a vertical tangent line + + {{hide|Consider the function $f(x)=x^{1 \over\ 3}$. Plotting the graph: [[Image: math images 7.jpg|left|thumb]] + + If you take the $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {a^{1 \over\ 3} \over\ a} = \lim_{a \to 0} {1 \over\ a^{2 \over\ 3}}$ + + As $a$ approaches infinity, the denominator becomes smaller and smaller, so the function grows beyond bounds, hence no derivative at $x=0$. The function is therefore not differentiable.}} + + + + + '''Note: While all differentiable functions are continuous, all continuous functions may not be differentiable.''' + |Pre-K=No + |Elementary=No + |MiddleSchool=No + |HighSchool=No + |AuthorName=Lizah Masis + |SiteName=made with Mathematica + |SiteURL=http://wikis.swarthmore.edu/miwiki/index.php/User:Lmasis1 + |Field=Calculus + |FieldLinks=Visit this site for an interactive experience with differentiability: + :*http://www-math.mit.edu/18.013A/HTML/chapter06/section01.html + } + |Field=Algebra + |InProgress=No + }} {{HelperPage|1=Roulette|2=Strange Attractors|3=Critical Points|4=Taylor Series}} {{HelperPage|1=Roulette|2=Strange Attractors|3=Critical Points|4=Taylor Series}} {{Image Description {{Image Description

## Revision as of 12:49, 2 July 2013

A Differentiable Function
$y=x^2$ is an example of a function that is differentiable everywhere.

# Basic Description

If we were to draw a tangent through every point on the curve in the main image, we would not encounter any difficulty at any point because there are no discontinuities, sharp corners and straight vertical portions at any point. This means that the function is differentiable.

# A More Mathematical Explanation

A function is differentiable at a point if it has a tangent at every point. That is, a function is di [...]

A function is differentiable at a point if it has a tangent at every point. That is, a function is differentiable at $x$ if the limit

$\lim_{a \to 0} {{f(x+a)-f(x)} \over\ a}$ exists.

It fails to be differentiable if:

• $f(x)$ is not continuous at $a$

The limit does not approach the same value from the right as it does from the left, so the function is not differentiable

For example $f(x)= {1 \over\ x}$ is not continuous at $x=0$. The function is undefined at that point, hence it is not differentiable. Computing the limit:

$\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {{{1 \over\ a} -0} \over\ a}$ $= \lim_{a \to 0} {1 \over\ a^2}$

As $a$ approaches $0$ from the left and right,the denominator becomes smaller and smaller, hence the limit approaches $- \infty$ and $\infty$ respectively. The limits from the right and left are different so the limit does not exist hence the function is not differentiable.

• The graph has a sharp corner at $a$

The function $f(x)=|x|$ has a sharp corner at $x=0$.

Computing the limit: $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {{|a|-0} \over\ a} = \lim_{a \to 0} {{|a|} \over\ a}$

As $a$ approaches $0$ from the right, the ratio is $1$, from the left, the ratio is $-1$. The limits are different, so the function is not differentiable.
• The graph has a vertical tangent line

Consider the function $f(x)=x^{1 \over\ 3}$. Plotting the graph:

If you take the $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {a^{1 \over\ 3} \over\ a} = \lim_{a \to 0} {1 \over\ a^{2 \over\ 3}}$

As $a$ approaches infinity, the denominator becomes smaller and smaller, so the function grows beyond bounds, hence no derivative at $x=0$. The function is therefore not differentiable.

Note: While all differentiable functions are continuous, all continuous functions may not be differentiable.

# Teaching Materials

Visit this site for an interactive experience with differentiability:

}

This is a Helper Page for:
Roulette
Strange Attractors
Critical Points
Taylor Series

A Differentiable Function
$y=x^2$ is an example of a function that is differentiable everywhere.

# Basic Description

If we were to draw a tangent through every point on the curve in the main image, we would not ecounter any difficulty at any point because there are no discontinuities, sharp corners and straight vertical portions at any point. This means that the function is differentiable.

# A More Mathematical Explanation

A function is differentiable at a point if it has a tangent at every point. That is, a function is di [...]

A function is differentiable at a point if it has a tangent at every point. That is, a function is differentiable at $x$ if the limit

$\lim_{a \to 0} {{f(x+a)-f(x)} \over\ a}$ exists.

It fails to be differentiable if:

• $f(x)$ is not continuous at $a$

The limit does not approach the same value from the right as it does from the left, so the function is not differentiable

For example $f(x)= {1 \over\ x}$ is not continuous at $x=0$. The function is undefined at that point, hence it is not differentiable. Computing the limit:

$\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {{{1 \over\ a} -0} \over\ a}$ $= \lim_{a \to 0} {1 \over\ a^2}$

As $a$ approaches $0$ from the left and right,the denominator becomes smaller and smaller, hence the limit approaches $- \infty$ and $\infty$ respectively. The limits from the right and left are different so the limit does not exist hence the function is not differentiable.

• The graph has a sharp corner at $a$

The function $f(x)=|x|$ has a sharp corner at $x=0$.

Computing the limit: $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {{|a|-0} \over\ a} = \lim_{a \to 0} {{|a|} \over\ a}$

As $a$ approaches $0$ from the right, the ratio is $1$, from the left, the ratio is $-1$. The limits are different, so the function is not differentiable.
• The graph has a vertical tangent line

Consider the function $f(x)=x^{1 \over\ 3}$. Plotting the graph:

If you take the $\lim_{a \to 0} {{f(a)-f(0)} \over\ a}= \lim_{a \to 0} {a^{1 \over\ 3} \over\ a} = \lim_{a \to 0} {1 \over\ a^{2 \over\ 3}}$

As $a$ approaches infinity, the denominator becomes smaller and smaller, so the function grows beyond bounds, hence no derivative at $x=0$. The function is therefore not differentiable.

Note: While all differentiable functions are continuous, all continuous functions may not be differentiable.

# Teaching Materials

Visit this site for an interactive experience with differentiability: