`star` Differentiability

`star` Theorem : If A function f is differentiable at a point c, then it is also continuous at that point.

`star` Derivatives of composite functions

`star` Theorem : (Chain Rule)

`star` Theorem : If A function f is differentiable at a point c, then it is also continuous at that point.

`star` Derivatives of composite functions

`star` Theorem : (Chain Rule)

● If `f` is a real function and `c` is a point in its domain. The derivative of `f` at `c` is defined by `color{red}{lim_(h-> 0) (f(c+h)-f(c))/h}`

●So a function `f` is differentiable at a point `c` in its domain if both `lim_(h-> 0^-) (f(c+h)-f(c))/h` and `lim_(h-> 0^+) (f(c+h)-f(c))/h` are finite and equal.

`color{blue}{lim_(h-> 0^-) (f(c+h)-f(c))/h = lim_(h-> 0^+) (f(c+h)-f(c))/h}`

●Derivative of `f` at `c` is denoted by `f '(c)` or `d/(dx) (f(x))|_c` The function defined by

`f'(x) =lim_(h->0) (f(x+h)-f(x))/h`- The derivative of `f` is denoted by `f ' (x)` or `d/(dx) (f(x))` or `y'` `f (x)` with respect to `x` to mean find `f '(x)`.

The following rules of algebra of derivatives :

(1) `(u pm v)' = u' pm v'`

(2) `(uv)'= u'v + uv'` (Leibnitz or product rule)

(3) `(u/v)' =((u'v -uv')/v^2)` wherever `v != 0` (Quotient rule).

The following table gives a list of derivatives of certain standard functions:

`f(x)` | `x^n` | `sin x` | `cos x` | `tan x` |

`f'(x)` | `nx^(n-1)` | `cos x` | `-sin x` | `sec^2 x` |

Differentiation formulas list :

`\color{green} ✍️` Since `f` is differentiable at `c`, we have

`lim_(x-> c) (f(x)-f(c))/(x-c)= f'(c)`

● But for `x ≠ c`, we have

`f(x) -f(c) =(f(x) -f(c))/(x-c) * (x-c)`

Therefore `lim_(x->c) [ f(x) -f(c)] = lim_(x-> c) [(f(x)-f(c))/(x-c) * (x-c)]`

or `lim_(x-> c) [f(x)]- lim_(x-> c) [f(c)] `

`=lim_(x-> c) [(f(x)-f(c))/(x-c)] * lim_(x-> c) [(x-c)]`

`= f'(c) * 0=0`

or `lim_(x-> c) f(x) = f(c)`

Hence `f` is continuous at `x = c`.

`color{red}{"Note : So Every differentiable function is continuous at that point but"}`

`color{red}{" that doesn't mean Every continuous function is differentiable at the same point."}`

E.g. Check differentiablity and continuity of `f(x) = |x|`

Consider the left hand limit

`=>lim_(x-> 0^(-)) f(x) =lim_(x-> 0^(-)) (-x) =0 `

`=>= lim_(x->0^+) f(x) =lim_(x-> 0^+) x=0`

● So `|x|` is continuous at `x=0`

Now for differentiablity

LHD `lim_(h-> 0^(-)) (f(0+h) -f(0))/h =(-h)/h=-1`

RHD `lim_(h-> 0^(+)) (f(0+h) -f(0))/h =h/h =1`

● Since the above left and right hand limits at `0` are not equal `lim_(h-> 0) (f(0+h) -f(0))/h` does not exist

and hence f is not differentiable at `0`. Thus `f` is not a differentiable function.

● So `|x|` is continuous at `x=0` but not differentiable.

`lim_(x-> c) (f(x)-f(c))/(x-c)= f'(c)`

● But for `x ≠ c`, we have

`f(x) -f(c) =(f(x) -f(c))/(x-c) * (x-c)`

Therefore `lim_(x->c) [ f(x) -f(c)] = lim_(x-> c) [(f(x)-f(c))/(x-c) * (x-c)]`

or `lim_(x-> c) [f(x)]- lim_(x-> c) [f(c)] `

`=lim_(x-> c) [(f(x)-f(c))/(x-c)] * lim_(x-> c) [(x-c)]`

`= f'(c) * 0=0`

or `lim_(x-> c) f(x) = f(c)`

Hence `f` is continuous at `x = c`.

`color{red}{"Note : So Every differentiable function is continuous at that point but"}`

`color{red}{" that doesn't mean Every continuous function is differentiable at the same point."}`

E.g. Check differentiablity and continuity of `f(x) = |x|`

Consider the left hand limit

`=>lim_(x-> 0^(-)) f(x) =lim_(x-> 0^(-)) (-x) =0 `

`=>= lim_(x->0^+) f(x) =lim_(x-> 0^+) x=0`

● So `|x|` is continuous at `x=0`

Now for differentiablity

LHD `lim_(h-> 0^(-)) (f(0+h) -f(0))/h =(-h)/h=-1`

RHD `lim_(h-> 0^(+)) (f(0+h) -f(0))/h =h/h =1`

● Since the above left and right hand limits at `0` are not equal `lim_(h-> 0) (f(0+h) -f(0))/h` does not exist

and hence f is not differentiable at `0`. Thus `f` is not a differentiable function.

● So `|x|` is continuous at `x=0` but not differentiable.

● The derivative of `f`, where `f (x) = (2x + 1)^3`

`=>` One way is to expand `(2x + 1)^3` using binomial theorem and find the derivative as a polynomial function as illustrated below.

`d/(dx) f(x) =d/(dx) [(2x+1)^3]`

`=d/(dx) (8x^3+12x^2+6x+1)`

`= 24x^2 + 24x + 6`

`= 6 (2x + 1)^2`

Now, observe that `f (x) = (h o g) (x)`

where `g(x) = 2x + 1` and `h(x) = x^3`.

Put `t = g(x) = 2x + 1`. Then `f(x) = h(t) = t^3`.

Thus `(df)/(dx) =6(2x+1)^2 =3(2x+1)^2 * 2 =3t^2 * 2 =(dh)/(dt) * (dt)/(dx)`

● The advantage with such observation is that it simplifies the calculation in finding the derivative of, say, `(2x + 1)^100`.

● We may formalise this observation in the theorem called the chain rule explained in next block.

`=>` One way is to expand `(2x + 1)^3` using binomial theorem and find the derivative as a polynomial function as illustrated below.

`d/(dx) f(x) =d/(dx) [(2x+1)^3]`

`=d/(dx) (8x^3+12x^2+6x+1)`

`= 24x^2 + 24x + 6`

`= 6 (2x + 1)^2`

Now, observe that `f (x) = (h o g) (x)`

where `g(x) = 2x + 1` and `h(x) = x^3`.

Put `t = g(x) = 2x + 1`. Then `f(x) = h(t) = t^3`.

Thus `(df)/(dx) =6(2x+1)^2 =3(2x+1)^2 * 2 =3t^2 * 2 =(dh)/(dt) * (dt)/(dx)`

● The advantage with such observation is that it simplifies the calculation in finding the derivative of, say, `(2x + 1)^100`.

● We may formalise this observation in the theorem called the chain rule explained in next block.

`\color{green} ✍️` (i) If `f` be a real valued function which is a composite of two functions `u` and `v` ; i.e.,` f = v o u`. Suppose `t = u(x)` and if both `(dt)/(dx)` and `(dv)/(dt)` exist,

we have `color{orange}{(df)/(dx) =(dv)/(dt) * (dt)/(dx)}`

Suppose `f` is a real valued function which is a composite of three functions `u, v` and `w` ; i.e.,

`\color{green} ✍️` (ii) `f = (w o u) o v`. If `t = v (x)` and `s = u (t)`, then

`color{orange}{(df)/(dx) =(d (wou))/(dt) * (dt)/(dx) =(dw)/(ds) * (ds)/(dt) * (dt)/(dx)}`

provided all the derivatives in the statement exist.

E.g., : Differentiate `sin (cos (x^2))` with respect to `x`.

The function `f (x) = sin (cos (x^2))` is a composition `f (x) = (w o v o u) (x)` of the three functions `u, v` and `w`,

where `u(x) = x^2, v(t) = cos t` and `w(s) = sin t`.

Put `t = u(x) = x^2` and `s = v(t) = cos t`. Observe that `(dw)/(ds) =cos s, (ds)/(dt) =-sin t` and `(dt)/(dx) =2x` exist for all real `x`.

Hence by a generalisation of chain rule, we have

`(df)/(dx) =(dw)/(ds) * (ds)/(dt) * (dt)/(dx) =(cos s) * (-sin t) * (2x) =-2x sin x^2 * cos (cos x^2)`

we have `color{orange}{(df)/(dx) =(dv)/(dt) * (dt)/(dx)}`

Suppose `f` is a real valued function which is a composite of three functions `u, v` and `w` ; i.e.,

`\color{green} ✍️` (ii) `f = (w o u) o v`. If `t = v (x)` and `s = u (t)`, then

`color{orange}{(df)/(dx) =(d (wou))/(dt) * (dt)/(dx) =(dw)/(ds) * (ds)/(dt) * (dt)/(dx)}`

provided all the derivatives in the statement exist.

E.g., : Differentiate `sin (cos (x^2))` with respect to `x`.

The function `f (x) = sin (cos (x^2))` is a composition `f (x) = (w o v o u) (x)` of the three functions `u, v` and `w`,

where `u(x) = x^2, v(t) = cos t` and `w(s) = sin t`.

Put `t = u(x) = x^2` and `s = v(t) = cos t`. Observe that `(dw)/(ds) =cos s, (ds)/(dt) =-sin t` and `(dt)/(dx) =2x` exist for all real `x`.

Hence by a generalisation of chain rule, we have

`(df)/(dx) =(dw)/(ds) * (ds)/(dt) * (dt)/(dx) =(cos s) * (-sin t) * (2x) =-2x sin x^2 * cos (cos x^2)`

Q 3115323269

Find the derivative of the function given by `f (x) = sin (x^2)` .

Class 12 Chapter 5 Example 21

Class 12 Chapter 5 Example 21

Observe that the given function is a composite of two functions. Indeed, if

`t = u(x) = x^2` and `v(t) = sin t`, then

`f (x) = (v o u) (x) = v(u(x)) = v(x^2) = sin x^2`

Put t =` u(x) = x^2`. Observe that ` (dv)/(dt) = cos t` and ` (dt)/(dx) = 2x` exist. Hence, by chain rule

`(df)/(dx) = (dv)/(dt) * (dt)/(dx) = cos t * 2x`

It is normal practice to express the final result only in terms of x. Thus

` (df)/(dx) = cos t * 2x = 2x cos x^2`

Alternatively, We can also directly proceed as follows:

`y = sin (x^2) => (dy)/(dx) = d/(dx) (sin x^2 )`

`= cos x^2 d/(dx) (x^2) = 2x cos x^2`

Q 3145423363

Find the derivative of tan (2x + 3).

Class 12 Chapter 5 Example 22

Class 12 Chapter 5 Example 22

Let f (x) = tan (2x + 3), u (x) = 2x + 3 and v(t) = tan t. Then

(v o u) (x) = v(u(x)) = v(2x + 3) = tan (2x + 3) = f (x)

Thus f is a composite of two functions. Put t = u(x) = 2x + 3. Then `(dv)/(dt) = sec^2 t` and

`(dt)/(dx) = 2 ` exist Hence, by chain rule

` (df)/(dx) = (dv)/(dt) * (dt)/( dx) = 2 sec^2 (2x + 3)`

Q 3165423365

Differentiate `sin (cos (x^2))` with respect to x.

Class 12 Chapter 5 Example 23

Class 12 Chapter 5 Example 23

The function `f (x) = sin (cos (x^2))` is a composition f (x) = (w o v o u) (x) of the

three functions u, v and w, where `u(x) = x^2, v(t) = cos t` and w(s) = sin s. Put

`t = u(x) = x^2` and `s = v(t) = cos t` . Observe that `(dw)/(ds) = cos s , (ds)/(dt) = -sin t` and ` (dt)/(dx) =2x`

exist for all real x. Hence by a generalisation of chain rule, we have

` (df)/(dx) = (dw)/(ds) * (ds)/(dt) * (dt)/(dx) = (cos s) * ( - sin t) * (2x) = -2x sin x^2 * cos (cos x^2)`