Right hand & Left hand Derivatives : By definition: `f'(a) = Lim_(h->0) (f (a+h) -f (a))/h` if it exist

(a) The right hand derivative of `f` at `x = a` denoted by `f'(a+)` is defined by : `f' (a+) = Lim_(h->0^+) (f (a+h) -f (a))/h` ,
provided the limit exists & is finite.
when `h -> 0`, the point `B` moving along the curve tends to `A`, i.e., `B -> A` then the chord `AB` approaches
the tangent line AT at the point `A`.
`=> f'(a^+) = lim_(h->0) (f(a + h) - f(a))/h = lim_(h-> 0)tan phi =tan psi`

(b) The left hand derivative of fat `x = a` denoted by `f' (a^-)` is defined by - `f' (a^-)= Lim_(h->0^+) (f(a-h) -f (a))/-h`,


Provided the limit exists & is finite.
When `h -> 0`, the point `C` moving along the curve tends to `A`, i.e., `C-> A` then the chord `CA` approaches
the tangent line AT at the point `A` then

`=> f'(a^-) = lim_(h->0 (f(a-h) -f (a))/-h)`

`f` is said to be derivable at `x = a` if `f' (a^+ )= f' (a^- ) = a` finite quantity
This geomtrically means that a unique tangent with finite slope can be drawn at `x = a` as shown in the
figure

If a function f is derivable at `x =a` then `f` is continuous at `x =a`.

Proof:

Let `f`is derivable at `x = a`. Hence

For: `f'(a) = Lim_(h->0) (f(a+h) -f (a))/h` exist.

Also `f(a + h) - f(a) = ((f(a+h) -f (a)) /h )*h ` `( h ne 0)`

Therefore: `Lim_(h-> 0) [ ( f (a+h)) - f(a) ]= Lim_(h-> 0) ( ( f (a+h) -f (a) )/h)*h =f' (a) * 0 =0`

Therefore `Lim_(h->0) [ (f (a+h) -f (a)) ]= 0 => Lim_(h-> 0) f (a+h) = f (a) => f` is continuous at `x`.

Note: If `f(x)` is derivable for every point of its domain of definition, then it is continuous in that domain.

The Converse of the above result is not true :

For a function f :

Differentiability `=>` Continuity ; Continuity does not implies derivability ;

Non derivibality does not implies discontinuous But discontinuity => Non derivability


Note

(a) Let `f'(a^+) = p` & `f' (a) = q` where `p` & `q` are finite then:

(i) `p = q => f` is derivable at `x = a` `=>` `f` is continuous at `x = a`.
(ii) `p ne q => f` is not derivable at `x = a`.

It is very important to note that `f` may be still continuous at `x = a`.
(b) If a function `f` is not differentiable but is continuous at `x = a` it gcometrically implies a sharp comer
at `x = a`.

(a) The function `f(x)` is said to non-differentiable at `x = a` if
Both left and right and hand derivative exists but are not equal
The function `y = |x|` is not differentiable at `0` as its graph change
direction abrupty when `x = 0`. In general, if the graph of a function
has a 'comer' or 'kink' in it, then the graph of `f` has no tangent at this
point and `f` is not differentiable there. (To compute `f' (a)`, we find that
the left and tight limits are different.)

(b) Function is discontinuous at `x =a`

If `f` is not continuous at `a` then `f` is not differentiable at `a`. So at any
discontinuity (for instance, `a` jump of discontinuity) `f` fai ls to be
differentiable

(c) Either or both left and tight hand derivative are not finite.

A third possibility is that the curve has a vertical tangent line when
`x = a`, that is `f` is continuous at `a` and `lim_(x->a) |f' (x) | = oo`

This means that the tangent lines becomes steeper and sleeper
as `x -> a`.

`f(x)` is said to be derivable over an open interval `(a, b)` if it is derivable at each & every point of the
interval `f(x)` is said to be derivable over the closed interval `[a, b]` if :

(i) for the points `a` and `b` , `f'(a^+)` & `f'(b^-)` exist &
(ii) for any point `c` such that `a < c < b,f'(c^+)` & `f(c^ - )` exist & are equal.

Note : Consider the graph of a differentiable function

(1) If `f(x)` is derivable at `x = x_0` then `| f(x) |` must be
derivable at `x = x_0` provided `f(x_0) ne 0`. However
if `f(x_0) =0` then `| f(x) |` mayor may not be
derivable at `x = 0`.

e.g. `f(x) = x^3` is derivable at `x = 0` and `| f(x) |` is also derivable at `x = 0`.
`f(x) = x - 1` is derivable at `x =1` but I f(x) I is not derivable at `x = 1`
i.e. if `f' (x_ 0) = 0` and `f(x_0) = 0` then `| f(x) |` will also derivable at `x = x_0` and if `f' (x _0)` is non
zero finite then `| f(x) |` is non derivable at `x= x_0`.


In figure `f(x) = 0` at `A, B, C` and `f(x)` is derivable at `A, B` and `C` but `| f(x) |` is non derivable at `x =B`
and `C` but derivable at `x = A`.
Consider `f(x) =x^3` at `x = 0` when `f'(0) = 0` and `f(0) = 0` but from `| f(x) |= | x^3 |` is derivable but if
`f(x) = x` , where `f' (x) ne 0`,
hence `| f(x) |` is not derivable at `x = 0`

lf `f(x)` and `g (x)` both are derivable at `x = a`, then
(i) `f(x) pm g(x) ` will be differentiable at `x = a`.
(ii) `f(x) *g(x) ` will be differentiable at `x = a`.
(iii) `f(x) /g(x)` w1ll be d1fferentmble at `x = a` Jf `g(a) ne 0`


Note that:

(1) lf `f(x)` and `g (x` )are both derivable at `x=a , f(x) pm g(x) ; g(x) * f(x)` and `f(x) /g(x)` will also bederivableat
`x = a`. (only if `g (a) ne 0`)

(2) lf `f(x)` is derivable at `x = a` and `g (x)` is not derivable at `x = a` then the `f(x) + g (x)` or `f(x)- g (x)` will not
be derivable at `x = a`.

e.g. `f(x) = cos | x |` is derivable at `x = 0` and `g (x) = | x |` is not derivable at `x = 0`,
then `cos | x | + | x |` or `cos | x | - | x |` will not be derivable at `x = 0`.
However nothing can be said about the product function in this case.

`f(x) = x` derivable at `x = 0`
`g (x) = | x |` not derivable at `x = 0`

`x | x | = [tt( (x^2, text (if) x ge 0) ,(-x^2, text(if) x < 0))`

(3) If both `f (x)` and `g (x)` are non derivable then nothing definite can be said about the
sum /difference / product function.

e.g. `f(x) = sin | x |` not derivable at `x = 0`
`g (x) = | x |` not derivable at `x = 0`
then the function
`F (x) = sin | x | - | x |` is derivable at `x = 0`
`G (x) = sin | x | + | x |` is not derivable at `x = 0`.



(4) lf `f(x)` is derivable at `x =a` and `f(a) = 0` and `g (x)` is continuous at `x = a` then the product function
`F (x) = f(x) * g (x)` will be derivable at `x = a`

`F' (a^+) =Lim_(h->0) ( f (a=h) g (a+h) -0)/h= f' (a) *g(a)`

`F' (a^-) = Lim_(h->0) ( f(a-h) g (a-h) -0 )/-h = f '(a) *g(a)`



(5) Derivative of a continuous function need not be a continuous function

`f(x) = {tt( ( x^2sin(1/x) , text(if) x ne 0 f'(0^+) =0) , (0, text (if) x=0 ; f' (0^-) =0) )`


`f(x) = {tt( (sin(1/x) * 2x-x^2 cos (1/x) (-1/x^2) , x ne 0) , ( 0 , text(if) x=0) )`


`f' (x)` is not continuous at `x = 0`.

Basic Steps:

(i) Write down the expression for `f' (x)` as `f' (x) = ( f (x+h) -f (x))/h`

(ii) Manipulate `f(x + h)- f(x)` in such a way that the given functional rule is applicable. Now apply the
functional rule and simplify the `RHS` to get `f' (x)` as a function of `x` along with constants if any.
(iii) Integrate `f' (x)` get `f(x)` as a function of `x` and a constant of integration. In some cases a Differential
Equation in formed which can be solved to get `f(x)`.
(iv)Apply the boundary value conditions to determine the value of this constant.