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Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

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Differentiability of Functions

» A function `f(x)` is differentiable at point `x=a` *if the "derivative-limit" is defined* at that point.

→ right-hand-limit `lim_(x->a+) (f(x)-f(a))/(x-a)`

→ left-hand-limit `lim_(x->a-) (f(x)-f(a))/(x-a)`

→ if both the limits are equal

→ and if both the limits are real numbers

→ eg: `f(x)=|x|` is not differentiable at `x=0`

*plain and simple summary*

nub

*plain and simple summary*

nub

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Continuity of "derivative-limit" is the differentiability of the function.

*simple steps to build the foundation*

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*simple steps to build the foundation*

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In this page, the conditions under which derivative is defined for functions is discussed.

Stating on learning "Differentiability of Functions". In this page, the conditions under which derivative is defined for functions is discussed.

Let us revise what we learned in "limits"

• when the function evaluates to indeterminate value `0//0`, limit of the function helps to resolve its value.

• The function is defined at a point, if left-hand-limit and right-hand-limit are equal at the given point.

• The function is not defined at a point, if left-hand-limit and right-hand-limit are not equal at the given point.

• When, left-hand-limit and right-hand-limit are equal, it is commonly referred as limit of the function.

The derivative of a function `f(x)` is the instantaneous rate of change given as

`lim_(delta ->0) (f(x+delta) - f(x))/(delta)`

Are left-hand-limit and right-hand-limit applicable to this expression?

- yes. For some functions, LHL and RHL are to be computed to establish that this limit exists
- No. For rate of change, the limit always exists.

The answer is "yes. For some functions, LHL and RHL are to be computed to establish that this limit exists". The objective of this lesson is to learn when the LHL and RHL are computed.

The figure shows `y=|x|` in blue. We need to compute `(dy)/(dx)`. `y= { (x , text( for ) x>0), (-x , text( for ) x <0), (0 , text( for ) x =0):}`

Derivative by first principles

`(dy)/(dx)|_(x=a)`*for `x>0` : `|x|=x` and `|a|=a`**right-hand-limit*

`= lim_(x->a+delta) (f(x)-f(a))/(x-a)`

`= lim_(x->a+delta) (|x|-|a|)/(x-a)`

`= lim_(x->a+delta) (x-a)/(x-a)`

`= (a+delta-a)/(a+delta-a)`

`= (delta)/delta`

`= 1`*left-hand-limit*

`= lim_(x->a-delta) (f(x)-f(a))/(x-a)`

`= lim_(x->a-delta) (|x|-|a|)/(x-a)`

`= lim_(x->a-delta) (x-a)/(x-a)`

`= (a-delta-a)/(a-delta-a)`

`= (-delta)/(-delta)`

`= 1`

What does this imply for `x>0`?

- the limit exists for `x>0`
- the expression under limit is continuous for `x>0`
- both the above

The answer is "both the above". Note that, the expression under limit is `(f(x+delta) -f(x))/delta`.

The figure shows `y=|x|` in blue. Summary of the result from previous page.

`(dy)/(dx)|_(x=a)`*for `x>0` : `|x|=x` and `|a|=a`**right-hand-limit*

`= lim_(x->a+delta) (|x|-|a|)/(x-a)`

`= 1`*left-hand-limit*

`= lim_(x->a-delta) (|x|-|a|)/(x-a)`

`= 1`

Note that, the expression under limit is `(f(x+delta) -f(x))/delta`. This expression is continuous and defined for x>0.

Since the expression is the derivative of the function `f(x)`, *the derivative of the function is defined for `x>0`*.

The figure shows `y=|x|` in blue. We need to compute `(dy)/(dx)`. Derivative by first principles:

`(dy)/(dx)|_(x=a)`*for `x<0` : `|x|=-x` and `|a|=-a`**right-hand-limit*

`= lim_(x->a+delta) (f(x)-f(a))/(x-a)`

`= lim_(x->a+delta) (|x|-|a|)/(x-a)`

`= lim_(x->a+delta) (-x-(-a))/(x-a)`

`= (-a-delta+a)/(a+delta-a)`

`= (-delta)/delta`

`= -1`*left-hand-limit*

`= lim_(x->a-delta) (f(x)-f(a))/(x-a)`

`= lim_(x->a-delta) (|x|-|a|)/(x-a)`

`= lim_(x->a-delta) (-x-(-a))/(x-a)`

`= (-a+delta+a)/(a-delta-a)`

`= (delta)/(-delta)`

`= -1`

What does this imply for `x<0`?

- the limit exists for `x<0`
- the expression under limit is continuous for `x<0`
- both the above

The answer is "both the above". Note that, the expression under limit is `(f(x+delta) -f(x))/delta`.

The figure shows `y=|x|` in blue. Summary of the result from previous page.

`(dy)/(dx)|_(x=a)`*for `x<0` : `|x|=-x` and `|a|=-a`**right-hand-limit*

`= lim_(x->a+delta) (|x|-|a|)/(x-a)`

`= -1`*left-hand-limit*

`= lim_(x->a-delta) (|x|-|a|)/(x-a)`

`= -1`

Note that, the expression under limit is `(f(x+delta) -f(x))/delta`. This expression is continuous and defined for `x<0`.

Since the expression is the derivative of the function `f(x)`, *the derivative of the function is defined for `x<0`*.

The figure shows `y=|x|` in blue. We need to compute `(dy)/(dx)`. Derivative by first principles: finding limits at `x=0`

`y= | [x , text( for ) x>0], [-x , text ( for ) x <0], [0 , text ( for ) x =0]|`

`(dy)/(dx)|_(x=0)`*left hand limit*

`= lim_(x->0-delta) (f(x)-f(0))/(x-0)`

`= lim_(x->0-delta) (|x|-|0|)/(x-0)`*for `x=0-delta` : `|x|=delta`*

`= (delta-0)/(0-delta-0)`

`= (delta)/(-delta)`

`= -1`*right hand limit*

`= lim_(x->0+delta) (f(x)-f(0))/(x-0)`

`= lim_(x->0+delta) (|x|-|0|)/(x-0)`*for `x=0+delta` : `|x|=delta`*

`= (delta-0)/(0+delta-0)`

`= (delta)/(delta)`

`= 1`

At `x=0`, left-hand-limit is `-1` and right-hand-limit is `1`.

What does this imply?

- the limit does not exist at `x=0`
- the expression under limit is not continuous at `x=0`
- both the above

The answer is "both the above". Note that, the expression under limit is `(f(x+delta) -f(x))/delta`. This expression is not continuous at `x=0`, i.e. the limit does not exist.

The figure shows `y=|x|` in blue. `(dy)/(dx)|_(x=0)`*left-hand-limit*

`= lim_(x->0-delta) (|x|-|0|)/(x-0)`

`= -1`*right-hand-limit*

`= lim_(x->0+delta) (|x|-|0|)/(x-0)`

`= 1`

Note that, the expression under limit is `(f(x+delta) -f(x))/delta`. This expression is not defined at `x=0`. Since the expression is the derivative of the function `f(x)`, the derivative of the function is not defined for `x=0`.

The figure shows `y=|x|` in blue. It is proven that

• for `x<0`, the derivative is defined.

• for `x>0`, the derivative is defined.

• for `x=0`, the derivative is not defined.

This is referred as *differentiability of the function. *

• *The function is not differentiable at `x=0`*.

• *The function is differentiable for `x!=0`*.

Finding differentiability of `f(x) = tan x`.

Derivative is `(d)/(dx) tan x = sec^2 x`.*the result of derivative is proven later*

Figure shows `tan x` in blue and `sec^2 x` in orange. From the left and right side limits of `tan x`, it is found that the function is not continuous at `x=pi/2 + n pi`.

At `x=pi/2 + n pi` where `n=0,1,2, cdots`,

the derivative

`= sec^2 (pi/2)`

`= oo`

Is `tan x` a continuously differentiable function?

- No. `tan x` is not differentiable at `x=pi/2 + n pi`
- Yes. `tan x` is differentiable.

The answer is "No. `tan x` is not differentiable at `x=pi/2 + n pi`". It can be observed in the figure that the derivative `sec^2 x` takes very high value.

Finding differentiability of `f(x) = x^(1/3)`.

Derivative is `(d)/(dx) f(x) = x^(-2/3)`.*the result of derivative is proven later*

Figure shows `x^(1/3)` in blue and `x^(-2/3)` in orange. From the left and right side limits of `x^(1/3)`, the function is continuous at `x=0`.

At `x=0`, the derivative

`= 0^(-2/3)`

`= oo`

Is `x^(1//3)` a continuously differentiable function?

- No. It is not differentiable at `x=0`
- Yes. `x^(1//3)` is differentiable.

The answer is "No. It is not differentiable at `x=0`". It can be observed in the figure that the derivative `x^(-2//3)` takes very high value at `x=0`.

Finding differentiability of `f(x) = x^(1/2)`.

Derivative is `(d)/(dx) f(x) = x^(-1/2)`.*the result of derivative is proven later*

Figure shows `x^(1/2)` in blue and `x^(-1/2)` in orange. Note that the function is not defined for `x<0`.

Is `x^(1/2)` a continuously differentiable function?

- No. It is not differentiable for `x<0`
- Yes. `x^(1/2)` is differentiable

The answer is "No. It is not differentiable for `x<0`".

*comprehensive information for quick review*

Jogger

*comprehensive information for quick review*

Jogger

dummy

**Differentiability of a Function: ** A function `f(x)` is differentiable at point `x=a` if the limit is defined. That is,

right-hand-limit `lim_(x->a+) (f(x)-f(a))/(x-a)` and left hand limit `lim_(x->a-) (f(x)-f(a))/(x-a)` are equal and evaluates to a real value.

*practice questions to master the knowledge*

Exercise

*practice questions to master the knowledge*

Exercise

Finding differentiability of `f(x) = x^(-1)`

Derivative is `(d)/(dx) f(x) = x^(-2)`.*the result of derivative is proven later*

Figure shows `x^(-1)` in blue and `x^(-2)` in orange. From the left and right side limits of `x^(-1)`, it is derived that the function is not continuous at `x=0`.

At `x=0`, the derivative

`= 0^(-2)`

`= oo`

Is `x^(-1)` a continuously differentiable function?

- No. It is not differentiable at `x=0`
- Yes. `x^(-1)` is differentiable.

The answer is "No. It is not differentiable at `x=0`". It can be observed in the figure that the derivative `x^(-2)` takes very high value at `x=0`.

Finding differentiability of `f(x) = x^(-2)`.

Derivative is `(d)/(dx) f(x) = x^(-3)`.*the result of derivative is proven later*

Figure shows `x^(-2)` in blue and `x^(-3)` in orange. From the left and right side limits of `x^(-2)`, it is derived that the function is not continuous at `x=0`.

At `x=0`, the derivative

`= d/(dx) x^(-2)|_(x=0)`*from right and left hand limits *

`= +-oo`

Is `x^(-2)` a continuously differentiable function?

- No. It is not differentiable at `x=0`
- Yes. `x^(-2)` is differentiable.

The answer is "No. It is not differentiable at `x=0`". It can be observed in the figure that the derivative `x^(-3)` takes `+-oo` at `x=0`.

*your progress details*

Progress

*About you*

Progress

Let us revise what we learned in "limits" ;; when the function evaluates to indeterminate value 0 by 0, limit of the function helps to resolve its value. ;; The function is defined at a point, if left-hand-limit and right-hand-limit are equal at the given point. ;; The function is not defined at a point if left-hand-limit and right-hand-limit are not equal at the given point. ;; When, left-hand-limit and right-hand-limit are equal, it is commonly referred as limit of the function. ;; The derivative of a function f of x is the instantaneous rate of change given as limit delta tending to 0, f of x plus delta, minus f of x by delta. ;; Are left-hand-limit and right-hand-limit applicable to this expression.

yes;s;computed;establish

yes. For some functions, LHL and RHL are to be computed to establish that this limit exists

no;always

No. For rate of change, the limit always exists.

The answer is "yes. For some functions, LHL and RHL are to be computed to establish that this limit exists". The objective of this lesson is to learn when the LHL and RHL are computed.

The figure shows y = mod x in blue. We need to compute d y by d x. ;; Derivative by first principles is given. This computes right and left hand limits for x greater than 0. ;; What does this imply for x greater than 0.

1

2

3

The answer is "both the above". Note that, the expression under limit is f of x + delta minus f of x, divided by delta.

The figure shows y = mod x in blue. Summary of the result from previous page is listed. For x greater than 0, the right and left hand limits are 1. ;; Note that, the expression under limit is derivative of f of x. This expression is continuous and defined for x greater than 0 . Since the expression is the derivative of the function f of x, the derivative of the function is defined for x greater than 0

The figure shows y = mod x in blue. We need to compute the derivative. ;; Derivative by first principles is given. This computes right and left hand limits for x less than 0. ;; What does this imply for x less than 0.

1

2

3

The answer is "both the above". Note that, the expression under limit is f of x + delta minus f of x, divided by delta.

The figure shows y = mod x in blue. Summary of the result from previous page is listed. For x less than 0, the right and left hand limits are negative 1. ;; Note that, the expression under limit is derivative of f of x. This expression is continuous and defined for x less than 0 . Since the expression is the derivative of the function f of x, the derivative of the function is defined for x less than 0

The figure shows y = mod x in blue. We need to compute the derivative. ;; Derivative by first principle at x=0 is given. At x = 0, the left hand limit is minus 1 and right hand limit is 1. What does this imply.

1

2

3

The answer is "both the above". Note that, the expression under limit is f of x + delta minus f of x, divided by delta. This expression is not continuous at x=0, that is the limit does not exist.

The figure shows y = mod x in blue. Summary of the result from previous page is listed. For x = 0, the left hand limit is negative 1 and right hand limit is 1. ;; Note that, the expression under limit is not defined at x = 0. Since the expression is the derivative of the function f of x, the derivative of the function is not defined for x = 0.

The figure shows y = mod x in blue. It is proven that for x less than 0, the derivative is defined. ;; For x greater than 0, the derivative is defined. ;; For x = 0, the derivative is not defined. ;; This is referred as differentiability of the function. The function is not differentiable at x=0. The function is differentiable for x not equals 0.

Finding differentiability of tan x. Derivative of tan x is secant squared x. Figure shows tan x in blue, and secant squared x in orange. From the left and right side limits of tan x, it is found that he function is not continuous at x = pi by 2 + n pi. ;; At x=pi by 2 + n pi, the derivative is infinity. ;; Is tan x a continuously differentiable function.

no;not;pi;2

No. tan x is not differentiable at x=pi by 2 + n pi

s;yes

Yes. tan x is differentiable.

The answer is "No. tan x is not differentiable at x=pi by 2 + n pi". It can be observed in the figure that the derivative takes very high value.

Finding differentiability of f of x = x power 1 by 3 . Derivative is x power minus 2 by 3. Figure shows the function in blue and the derivative in orange. From the left and right hand limits of the function, it is found that the function is continuous at x=0. ;; At x=0, the derivative is infinity. Is the function a continuously differentiable function.

no;not

No. It is not differentiable at x=0

s;yes

Yes. x^(1//3) is differentiable.

The answer is "No. It is not differentiable at x=0 "It can be observed in the figure that the derivative takes very high value at x=0

Finding differentiability of f of x = x power 1 by 2 . Derivative is x power minus 1 by 2. Figure shows the function in blue and the derivative in orange. Note that the function is not defined for x less than 0. Is the function a continuously differentiable function.

no;not

No. It is not differentiable for x less than 0

s;yes

yes. x power 1 by 2 is differentiable

The answer is "No. It is not differentiable for x less than 0"

Continuity of derivative limit is the differentiability of the function.

Differentiability of a Function: A function f of x is differentiable at point x=a if the limit is defined. That is, right hand limit and left hand limit are equal and evaluates to a real value.

Finding differentiability of f of x = x power minus 1 . Derivative is x power minus 2. Figure shows the function in blue and the derivative in orange. From the left and right hand limits of the function, it is derived that the function is not continuous at x=0. ;; At x=0, the derivative is infinity. Is the function a continuously differentiable function.

no;not

No. It is not differentiable at x=0

s;yes

Yes. x^(-1) is differentiable.

The answer is "No. It is not differentiable at x=0 "It can be observed in the figure that the derivative takes very high value at x=0

Finding differentiability of f of x = x power minus 2 . Derivative is x power minus 3. Figure shows the function in blue and the derivative in orange. From the left and right hand limits of the function, it is derived that the function is not continuous at x=0. ;; At x=0, the derivative is plus or minus infinity. Is the function a continuously differentiable function.

no;not

No. It is not differentiable at x=0

s;yes

Yes. x^(-2) is differentiable.

The answer is "No. It is not differentiable at x=0 "It can be observed in the figure that the derivative takes very high value at x=0