Please Wait... While Loading Full Video#### Class 11 Chapter 13 - LIMITS AND DERIVATIVES

`color{red} star` Derivatives - Definition

`color{red} star` Algebra of derivative of functions

`color{red} star` Derivative of `f(x) = x^n` is `nx^(n – 1)` for any positive integer `n`.

`color{red} star` Derivative of polynomial functions

`color{red} star` Algebra of derivative of functions

`color{red} star` Derivative of `f(x) = x^n` is `nx^(n – 1)` for any positive integer `n`.

`color{red} star` Derivative of polynomial functions

●`\color{fuchsia} {ul"Definition 1 : "}` Suppose `f` is a real valued function and `a` is a point in its domain of definition. The derivative of `f` at a is defined by

`color(blue)(lim_(h →0) ( f(a+h) - f(a))/h)` provided this limit exists.

Derivative of `f (x)` at `a` is denoted by `f′(a)`. Observe that `f′ (a)` quantifies the change in `f(x)` at a with respect to `x`.

`color(blue)(lim_(h →0) ( f(a+h) - f(a))/h)` provided this limit exists.

Derivative of `f (x)` at `a` is denoted by `f′(a)`. Observe that `f′ (a)` quantifies the change in `f(x)` at a with respect to `x`.

Q 3121612521

Find the derivative at `x = 2` of the function `f(x) = 3x.`

We have

`f'(2) = lim_(h→0) (f(2+h)-f(2))/(h)= lim_(h→0) (3(2+h)-3(2))/(h)`

`lim_(h→0) (6+3h-6)/(h) = lim_(h→0)(3h)/(h) = lim_(h→0) 3 = 3`

The derivative of the function `3x` at `x = 2` is `3.`

Q 3150812714

Find the derivative of the function `f(x) = 2x^2 + 3x – 5` at `x = –1.` Also prove that `f ′ (0) + 3f′ ( –1) = 0.`

We first find the derivatives of f(x) at `x = –1 `and at `x = 0.` We have

`f' (-1) = lim_(h→0) (f(-1+h)-f(-1))/h`

`= lim_(h→0) ([ 2(-1+h)^2+3(-1+h)-5]-[2(-1)^2+3(-1)-5])/h`

`= lim_(h→0) (2h^2-h)/h = lim_(h→0) (2h-1) = 2(0) -1 = -1`

and `f'(0) = lim_(h→0) (f(0+h)-f(0))/h`

` = lim_(h→0) ( [ 2(0+h)^2+3(0+h)-5] - [ 2(0)^2+3(0) -5])/h`

` = lim_(h→0) (2h^2+3h)/h = lim_(2h+3) = 2(0)+3 = 3`

`f'(0)+3f'(-1) = 0`

Q 3161612525

Find the derivative of `sin x` at `x = 0.`

Let `f(x) = sin x`. Then

`f'(0) = lim_(h→0) (f(0+h) -f(0))/(h)`

`= lim_(h→0) (sin(0+h) -sin (0)) /(h) = lim_(h→0) (sin h)/(h) = 1`

Q 3171612526

Find the derivative of `f(x) = 3 " at " x = 0` and at `x = 3.`

Since the derivative measures the change in function, intuitively it is clear that the derivative of the constant function must be zero at every point. This is indeed, supported by the following computation.

`f'(0) = lim_(h→0)(f(0+h)-f(0))/(h)= lim_(h→0)(3-3)/(h) = lim_(h→0)(0)/(h) = 0`

Similarly `f'(3) = lim_(h→0)(f(3+h) - f(3))/(h) lim_(h→0)(3-3)/(h)=0.`

We now present a geometric interpretation of derivative of a function at a point. Let `y = f(x)` be a function and let `P = (a, f(a))` and `Q = (a + h, f(a + h)` be two points close to each other on the graph of this function. The Fig 13.11 is now self explanatory.

We know that `f'(a) = lim_(h→0) (f(a+h)-f(a))/(h)`

From the triangle PQR, it is clear that the ratio whose limit we are taking is precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process, as h tends to 0, the point Q tends to P and we have

`= lim_(h→0) (f(a+h)-f(a))/(h) = = lim_(Q→P) (QR)/(PR)`

`= lim_(h→0) (f(a+h)-f(a))/(h)== lim_(Q→P) (QR)/(PR)`

This is equivalent to the fact that the chord PQ tends to the tangent at P of the curve `y = f(x). `Thus the limit turns out to be equal to the slope of the tangent. Hence

`f ′(a) = "tan" ψ .`

For a given function f we can find the derivative at every point. If the derivative exists at every point, it defines a new function called the derivative of f .

`\color{fuchsia} {ul"Definition 2 : "}` Suppose `f` is a real valued function, the function defined by

`color(blue)(lim_(h→0) ( f(x+h) - f(x))/h)`

wherever the limit exists is defined to be the derivative of `f` at `x` and is denoted by `f′(x)`.

`color(red)("This definition of derivative is also called the first principle of derivative.")`

Thus `color(blue)(f'(x) = lim_(h→0) (f (x+h) - f(x))/h)`

Clearly the domain of definition of `f′ (x)` is wherever the above limit exists.

There are `color(green)("different notations for derivative of a function.")`

`\color{green} ✍️ ` Sometimes `f′ (x)` is denoted by `color(blue)(d/(dx) ( f(x)))` or if `y = f(x)`. it is denoted by `color(blue)((dy)/(dx))`. This is referred to as derivative of `f(x)` or `y` with respect to `x`.

`\color{green} ✍️` It is also denoted by `color(blue)(D (f (x) ))`. Further, derivative of `f` at `x = a` is also denoted by `d/(dx) f(x) |_(a) ` or `(df)/(dx) |_(a)` or even `((df)/(dx))_(x= a)`

Q 3181612527

Find the derivative of `f(x) = 10x.`

Since `f′ ( x) =lim_(h→0) (f(x+h)-f(x))/(h)`

`lim_(h→0)(10(x+h)-10(x))/(h)`

`lim_(h→0) (10h)/(h) = lim_(h→0) (10)=10`

Q 3111612529

Find the derivative of `f(x) = x^2.`

We have, `f ′(x) lim_(h→0) (f(x+h)-f(x))/(h)`

`lim_(h→0)((x+h)^2-(x)^2)/(h) = lim_(h→0) (h+2x)=2x`

Q 3111712620

Find the derivative of the constant function `f (x) = a` for a fixed real number a.

We have, `f ′(x) =lim_(h→0) (f(x+h)-f(x))/(h)`

`lim_(h→0) (a-a)/(h) = lim_(h→0) (0)/(h) =0` "as" `h ≠ 0`

Q 3121712621

Find the derivative of `f(x) = 1/x`

We have `f ′(x) = lim_(h→0) (f(x+h)-f(x))/(h)`

`= lim_(h→0)(1/(x+h)-1/x)/h`

`= lim_(h→0)1/h[(x-(x+h))/(x(x+h))]`

`= lim_(h→0)1/h[(-h)/(x(x+h))] = = lim_(h→0) (-1)/(x(x+h)) = -1/(x^2)`

● Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules for derivatives to follow closely that of limits. We collect these in the following theorem.

● Let `f` and `g` be two functions such that their derivatives are defined in a common domain. Then

.

`color(red)((i))` Derivative of sum of two functions is sum of the derivatives of the functions.

`color(blue)[d/(dx) { f(x) + g(x) } = d/(dx) f(x) +d/(dx) g (x)]`.

`color(red)((ii))` Derivative of difference of two functions is difference of the derivatives of the functions.

`color(blue)[d/(dx) { f(x) - g(x) } = d/(dx) f(x) - d/(dx) g(x)]`.

`color(red)((iii))` Derivative of product of two functions is given by the following `color(red)"product rule."`

`color(blue)[d/(dx) { f(x) * g(x) } = d/(dx) f(x) * g(x) + f(x) * d/(dx) g(x)]`

`color(red)((iv))` Derivative of quotient of two functions is given by the following `color(red)"quotient rule"` (whenever the denominator is non–zero).

`color(blue)[d/(dx) ((f(x))/(g(x))) = ( d/(dx) f(x) * g(x) - f(x) d/(dx) g(x))/(g(x))^2]`

Proof : Let `color(red)(u = f (x) and v = g (x).)` Then

`color(green)((uv)′ = u′v + uv′)`

This is referred to a Leibnitz rule for differentiating product of functions or the product rule. Similarly, the quotient rule is

`color(orange)((u/v)' = (u'v-uv')/(v^2))`

Now, let us tackle derivatives of some standard functions. It is easy to see that the derivative of the function `f(x) = x` is the constant

function 1. This is because `color(blue)(f '(x)=lim_(h →0) (f(x+h)-f(x))/h = lim_(h →0)(x+h-x)/h`

`color(red)(lim_(h →0)1=1)`

We use this and the above theorem to compute the derivative of

`color(blue)(f(x) = 10x = x + .... + x)` (ten terms). By (i) of the above theorem

`color(red)((df)/(dx) = (d)/(dx) (x+....+x))` (ten terms)

`color(orange)(= d/(dx)x+..+ d/(dx) x)` (ten terms)

`color(navy)(= 1+ ... +1` (ten terms) `= 10.)`

● We note that this limit may be evaluated using product rule too. Write `f(x) = 10x = uv,` where u is the constant function taking value 10 everywhere and `v(x) = x.` Here, `f(x) = 10x = uv` we know that the derivative of u equals 0. Also derivative of `v(x) = x` equals 1. Thus by the product rule we have

`color(blue)(f ′(x) = (10x)′ = (uv)′ = u′v + uv′ = 0.x +10.1 =10)`

● On similar lines the derivative of `f(x) = x^2` may be evaluated. We have `f(x) = x^2 = x .x` and hence

`color(green)((df)/(dx) = d/(dx) (x.x) = d/(dx) (x).x+x.d/(dx)(x))`

`color(blue)(l=1x+x.1=2x`

More generally, we have the following theorem.

`color{blue} {"Theorem 6 :"}` Derivative of `f(x) = x^n` `is` `nx^(n – 1)` for any positive integer `n`.

`color(red)("Proof:")` By definition of the derivative function, we have

`color(green)(f'(x) = lim_(h →0) (f (x+h) - f(x))/h = lim_( h→0) (( x+h)^n-x^n)/h)`

Binomial theorem tells that `color(purple)((x+h)^n = ( text()^(n)C_0) x^n +( text()^(n)C_1) x^(n-1) h +...........+ ( text()^(n)C_n) h^n)`

and hence `color(purple)((x+h)^n - x^n = h( n x^(n-1) +........+ h^(n-1)) )` .

Thus `color(blue)(( df (x))/(dx) = lim_(h→0) (( x+h)^n - x^n)/h)`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lim_(h →0) ( h ( n x^(n-1) + .......+ h^(n-1)))/h`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lim_( h → 0) ( n x^(n-1) +........+ h^(n-1)) = color(red)(nx^(n-1))`

`color(green)"Alternatively,"` we may also prove this by induction on `n` and the `color(blue)"product rule"` as follows.

We have

`color(blue)(d/(dx) (x^n) = d/(dx) ( x . x^(n-1)))`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= d/(dx) (x) . (x^(n-1)) + x. d/(dx) (x^(n-1))` (by product rule)

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 1 . x^(n-1) + x . ( (n-1) x^(n-2))` (by induction hypothesis)

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x^(n-1) + (n-1) x^(n-1) = color(red)(n x^(n-1))`

Remark : The above theorem is true for all powers of x, i.e., n can be any real number (but we will not prove it here).

● Let `f` and `g` be two functions such that their derivatives are defined in a common domain. Then

.

`color(red)((i))` Derivative of sum of two functions is sum of the derivatives of the functions.

`color(blue)[d/(dx) { f(x) + g(x) } = d/(dx) f(x) +d/(dx) g (x)]`.

`color(red)((ii))` Derivative of difference of two functions is difference of the derivatives of the functions.

`color(blue)[d/(dx) { f(x) - g(x) } = d/(dx) f(x) - d/(dx) g(x)]`.

`color(red)((iii))` Derivative of product of two functions is given by the following `color(red)"product rule."`

`color(blue)[d/(dx) { f(x) * g(x) } = d/(dx) f(x) * g(x) + f(x) * d/(dx) g(x)]`

`color(red)((iv))` Derivative of quotient of two functions is given by the following `color(red)"quotient rule"` (whenever the denominator is non–zero).

`color(blue)[d/(dx) ((f(x))/(g(x))) = ( d/(dx) f(x) * g(x) - f(x) d/(dx) g(x))/(g(x))^2]`

Proof : Let `color(red)(u = f (x) and v = g (x).)` Then

`color(green)((uv)′ = u′v + uv′)`

This is referred to a Leibnitz rule for differentiating product of functions or the product rule. Similarly, the quotient rule is

`color(orange)((u/v)' = (u'v-uv')/(v^2))`

Now, let us tackle derivatives of some standard functions. It is easy to see that the derivative of the function `f(x) = x` is the constant

function 1. This is because `color(blue)(f '(x)=lim_(h →0) (f(x+h)-f(x))/h = lim_(h →0)(x+h-x)/h`

`color(red)(lim_(h →0)1=1)`

We use this and the above theorem to compute the derivative of

`color(blue)(f(x) = 10x = x + .... + x)` (ten terms). By (i) of the above theorem

`color(red)((df)/(dx) = (d)/(dx) (x+....+x))` (ten terms)

`color(orange)(= d/(dx)x+..+ d/(dx) x)` (ten terms)

`color(navy)(= 1+ ... +1` (ten terms) `= 10.)`

● We note that this limit may be evaluated using product rule too. Write `f(x) = 10x = uv,` where u is the constant function taking value 10 everywhere and `v(x) = x.` Here, `f(x) = 10x = uv` we know that the derivative of u equals 0. Also derivative of `v(x) = x` equals 1. Thus by the product rule we have

`color(blue)(f ′(x) = (10x)′ = (uv)′ = u′v + uv′ = 0.x +10.1 =10)`

● On similar lines the derivative of `f(x) = x^2` may be evaluated. We have `f(x) = x^2 = x .x` and hence

`color(green)((df)/(dx) = d/(dx) (x.x) = d/(dx) (x).x+x.d/(dx)(x))`

`color(blue)(l=1x+x.1=2x`

More generally, we have the following theorem.

`color{blue} {"Theorem 6 :"}` Derivative of `f(x) = x^n` `is` `nx^(n – 1)` for any positive integer `n`.

`color(red)("Proof:")` By definition of the derivative function, we have

`color(green)(f'(x) = lim_(h →0) (f (x+h) - f(x))/h = lim_( h→0) (( x+h)^n-x^n)/h)`

Binomial theorem tells that `color(purple)((x+h)^n = ( text()^(n)C_0) x^n +( text()^(n)C_1) x^(n-1) h +...........+ ( text()^(n)C_n) h^n)`

and hence `color(purple)((x+h)^n - x^n = h( n x^(n-1) +........+ h^(n-1)) )` .

Thus `color(blue)(( df (x))/(dx) = lim_(h→0) (( x+h)^n - x^n)/h)`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lim_(h →0) ( h ( n x^(n-1) + .......+ h^(n-1)))/h`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lim_( h → 0) ( n x^(n-1) +........+ h^(n-1)) = color(red)(nx^(n-1))`

`color(green)"Alternatively,"` we may also prove this by induction on `n` and the `color(blue)"product rule"` as follows.

We have

`color(blue)(d/(dx) (x^n) = d/(dx) ( x . x^(n-1)))`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= d/(dx) (x) . (x^(n-1)) + x. d/(dx) (x^(n-1))` (by product rule)

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 1 . x^(n-1) + x . ( (n-1) x^(n-2))` (by induction hypothesis)

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x^(n-1) + (n-1) x^(n-1) = color(red)(n x^(n-1))`

Remark : The above theorem is true for all powers of x, i.e., n can be any real number (but we will not prove it here).

We start with the following theorem which tells us the derivative of a polynomial function.

`color(red)(ul"Theorem : 7 ")` Let `f(x) = a_n x^n +a_(n-1) x^(n-1) +...........+ a_1 x+a_0` be a polynomial function, where `a_1` are all real numbers and `a_n ≠ 0.`

Then, the derivative function is given by

`color(blue)((df(x))/(dx) = n a_n x^(n-1) + (n-1) a_(n-1) x^(x-2) +.............+ 2a_2x+a_1)`

`color(red)(ul"Theorem : 7 ")` Let `f(x) = a_n x^n +a_(n-1) x^(n-1) +...........+ a_1 x+a_0` be a polynomial function, where `a_1` are all real numbers and `a_n ≠ 0.`

Then, the derivative function is given by

`color(blue)((df(x))/(dx) = n a_n x^(n-1) + (n-1) a_(n-1) x^(x-2) +.............+ 2a_2x+a_1)`

Q 3131712622

Compute the derivative of `6x^100 – x^55 + x.`

A direct application of the above theorem tells that the derivative of the above function is `600x^99 − 55x^54 +1.`

Q 3141712623

Find the derivative of `f(x) = 1 + x + x^2 + x^3 +... + x^50` at `x = 1.`

direct application of the above Theorem tells that the derivative of the above function is

`1 + 2x + 3x^2 + . . . + 50x^49. ` At `x = 1` the value of this function equals

`1 + 2(1) + 3(1)^2 + .. . + 50(1)^49 `

`= 1 + 2 + 3 + . . . + 50 = ((50)(51))/(2) = 1275.`

Q 3151712624

Find the derivative of `f(x) = (x+1)/(x)`

Clearly this function is defined everywhere except at `x = 0.`

We use the quotient rule with `u = x + 1` and `v = x.` Hence `u′ = 1` and `v′ = 1. `

Therefore `(df(x))/(dx) = d/(dx) (x+1)/(x) = d/(dx) (u/v) `

`" "= (u'v-uv')/(v^2) = (1(x) - ((x+1)1)) /(x^2) = -1/(x^2)`

Q 3161712625

Compute the derivative of `sin x.`

Let `f(x) = sin x.` Then

`(df(x))/(dx) = lim_(h→0) (f(x+h)-f(x))/(h)= lim_(h→0) (sin (x+h)- sin (x))/(h)`

`lim_(h→0) (2cos((2x+h)/(2)) sin (h/2))/(h)` (using formula for `sin A – sin B`)

`lim _(h→0) cos(x+h/2) lim_(h→0) (sin (h/2))/(h/2) = cos x. 1 = cos x`

Q 3171712626

Compute the derivative of tan `x.`

Let `f(x) =` tan `x.` Then

`(df(x))/(dx) = lim_(h→0) f(x+h)-f(x)/(h) = lim_(h→0) (tan(x+h)-tan (x)/(h)`

`lim_(h→0)1/h [ (sin(x+h))/(cos (x+h) - (sinx)/(cosx)]]`

`lim_(h→0)[ (sin (x+h) cosx - cos(x+h) sin x)/(hcos(x+h) cosx)]`

`lim_(h→0)(sin (x+h-x))/cos(x+h)cosx` (using formula for sin (A + B)

`lim_(h→0) (sinh)/(h) lim_(h→0) (1)/(cosx+h)cos x`

`= 1 (1)/(cos^2x) = sec^2x`

Q 3181712627

Compute the derivative of `f(x) = sin^2 x.`

We use the Leibnitz product rule to evaluate this.

`(df(x))/(dx) = d/(dx) (sinx sin x)`

`= (sin x)' sin x+sin x(sin x)'`

`= (cosx) sin x+ sin (cosx)`

`= 2sin x cosx= sin 2x.`