2.3 Differentiabilityap Calculus

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  4. 2.3 Differentiabilityap Calculus Definition

Differentiable means that the derivativeexists ..

Lesson 2.6: Differentiability: Afunctionisdifferentiable at a point if it has a derivative there. In other words: The function f is differentiable at x if.

Posted by 3 years ago. Can someone help me with this question? (Ap Calc AB - Graphing/differentiability) Ap calculus. Differentiate any function with our calculus solver. Enter an expression and the variable to differentiate with respect to. Then click the Differentiate button. FREE-RESPONSE SOLUTIONS 2019 AB Question AB-2 (a) vtP is differentiable and therefore continuous on 0.3,2.8.Since 2.8 0.3 55 55 0 2.8 0.3 2.5 vvPP. Its derivative is (1/3)x −(2/3) (by the Power Rule) At x=0 the derivative is undefined, so x (1/3) is not differentiable. At x=0 the function is not defined so it makes no sense to ask if they are differentiable there.

Example: is x2 + 6x differentiable?

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

2.3 differentiabilityap calculus calculator

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

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:

At x=0 it has a very pointy change!

Does the derivative exist at x=0?Format external hard drive for mac using windowslasopafs.

Testing

We can test any value 'c' by finding if the limit exists:

limh→0f(c+h) − f(c)h

Example (continued)

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

f(x) = x :limh→0 c+h − c h
Simplify:limh→0 h h

The limit does not exist! To see why, let's compare left and right side limits:

From Right Side:limh→0+ h h = +1

The limits are different on either side, so the limit does not exist.

So the function f(x) = x is not differentiable

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?

The absolute value function stays pointy even when zoomed in.

Other Reasons

Here are a few more examples:

The Floor and Ceiling Functions are not differentiable at integer values, as there is a discontinuity at each jump. But they are differentiable elsewhere.

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) is not differentiable.

At x=0 the function is not defined so it makes no sense to ask if they are differentiable there.

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


As we head towards x = 0 the function moves up and down faster and faster, so we cannot find a value it is 'heading towards'.

So it is not differentiable.

Different Domain

But we can change the domain!

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

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

2.3 Differentiabilityap Calculus Test

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, etc).

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.

(1) Find the derivatives of the following functions using first principle.

(i) f(x) = 6Solution

(ii) f(x) = -4x + 7Solution

(iii) f(x) = -x2 + 2 Solution

(2) Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

(i) f(x) = x - 1 Solution

(ii) f(x) = √(1 - x2) Solution

(3) Determine whether the following function is differentiable at the indicated values.

(i) f(x) = x x at x = 0 Solution

(ii) f(x) = x2 - 1 at x = 1 Solution

(iii) f(x) = x + x - 1 at x = 0, 1 Solution

(iv) f(x) = sin x at x = 0 Solution

(4) Show that the following functions are not differentiable at the indicated value of x.

(i)

Solution

(5) The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.

Solution

(6) If f(x) = x + 100 + x2, test whether f'(-100) exists.

(7) Examine the differentiability of functions in R by drawing the diagrams.

(i) sin x Solution

2.3 Differentiabilityap Calculus Notes

(ii) cos x Solution

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