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

*click on the content to continue..*

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
- 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
- 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
- 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
- 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`
- 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`
- 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`
- 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`".

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

**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.

*Solved Exercise Problem: *

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`
- 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`.

*Solved Exercise Problem: *

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`
- 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`.

*slide-show version coming soon*