# Differentiability

(Difference between revisions)
 Revision as of 13:49, 2 July 2013 (edit)m ← Previous diff Current revision (13:54, 2 July 2013) (edit) (undo)m Line 10: Line 10: It fails to be differentiable if: It fails to be differentiable if: - *$f(x)$ is not continuous at $a$ + *$f(x)$ 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]] {{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]] Line 26: Line 26: - *The graph has a sharp corner at $a$ + *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]] {{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]] Line 36: Line 36: 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.}} 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 + *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]] {{hide|Consider the function $f(x)=x^{1 \over\ 3}$. Plotting the graph: [[Image: math images 7.jpg|left|thumb]] Line 122: Line 122: |FieldLinks=Visit this site for an interactive experience with differentiability: |FieldLinks=Visit this site for an interactive experience with differentiability: :*http://www-math.mit.edu/18.013A/HTML/chapter06/section01.html :*http://www-math.mit.edu/18.013A/HTML/chapter06/section01.html + } + |Field=Algebra + |InProgress=No }} }}

## Current revision

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

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Roulette
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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

There are currently no teaching materials for this page. Add teaching materials.

Visit this site for an interactive experience with differentiability:

}

Have questions about the image or the explanations on this page?