Differentiable

Differentiable means that the derivative exists ...

Example: is x2 + 6x differentiable?

Derivative rules tell us the derivative of x2 is 2x and the derivative of x is 1, so:

Its derivative is 2x + 6

So yes! x2 + 6x is differentiable.

... and it must exist for every value in the function's domain.

Domain

In its simplest form the domain is
all the values that go into a function

Graph showing domain highlighted on the x-axis and range on the y-axis

Example (continued)

When not stated we assume that the domain is the Real Numbers.

For x2 + 6x, its derivative of 2x + 6 exists for all Real Numbers.

So we are still safe: x2 + 6x is differentiable.

But what about this:

Example: The function f(x) = |x| (absolute value):

|x| looks like this:   V-shaped graph of the absolute value function centered at the origin

At x=0 it has a very pointy change!

Does the derivative exist at x=0?

We often call these 'pointy' parts corners or cusps. When there's a sharp turn in a graph, the derivative likely doesn't exist there!

Testing

We can test any value "a" by finding if the limit exists:

lim h→0 f(a+h) − f(a)h

Example (continued)

Let's calculate the limit for |x| at the value 0:

Start with:lim h→0 f(a+h) − f(a)h f(x) = |x|:lim h→0 |a+h| − |a|h a=0:lim h→0 |h| − |0|h Simplify:lim h→0 |h|h

In fact that limit doesn't exist! To see why, let's compare left and right side limits:

From Left Side:lim h→0− |h|h = −1 From Right Side:lim h→0+ |h|h = +1

The limits are different on either side, so the limit doesn't exist at x=0

f(x) = |x| isn't differentiable at x=0

A good way to picture this in your mind is to think:

As I zoom in, does the function tend to become a straight line?

Comparison of a smooth curve becoming a line when zoomed and a corner remaining pointy

The absolute value function stays pointy at x=0 even when zoomed in.

Other Reasons

Here are a few more examples:

Step-like graph of the floor function showing jump discontinuities at integers  

The Floor and Ceiling Functions aren't differentiable at integer values, as there's a discontinuity at each jump. But they are differentiable elsewhere.

  
Cube root function graph showing a vertical tangent line at the origin  

The cube root function x(1/3)

Its derivative is (1/3)x-(2/3) (by the Power Rule)

At x=0 the derivative is undefined, so x(1/3) isn't differentiable, unless we exclude x=0.

1/x graph

  

At x=0 the function isn't defined so it makes no sense to ask if it is differentiable there.

To be differentiable at a certain point, the function must first of all be defined there!

Since 0 isn't in the domain of 1/x, we can still say 1/x is differentiable on its domain.

Graph of sine of one over x oscillating infinitely fast as it approaches zero		  

As we head toward x = 0 the function moves up and down faster and faster, so we can't find any value it is "heading toward".

So it isn't differentiable there.

Different Domain

But we can change the domain!

absolute positive graph

Example: The function g(x) = |x| with Domain (0, +∞)

The domain is from but not including 0 onwards (all positive values).

Which IS differentiable.

And I am "absolutely positive" about that :)

So the function g(x) = |x| with Domain (0, +∞) is differentiable.

We could also restrict the domain in other ways to avoid x=0 (such as all negative Real Numbers, all non-zero Real Numbers, and so on).

Why Bother?

Because when a function is differentiable we can use all the power of calculus when working with it.

Continuous

When a function is differentiable it is also continuous.

Differentiable ⇒ Continuous

But a function can be continuous but not differentiable. For example the absolute value function is actually continuous (though not differentiable) at x=0.

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Introduction to Derivatives Calculus Index