UNIT–4 | unit 4 differential calculus

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UNIT–4

Differential Calculus

4.1. Mean value theorem: Rolle’s theorem& Lagrange’s mean value theorem

Important definitions-

Continuity- suppose that a function f(x) is defined in the interval I , then it is said to be continuous at x=a , if

Differentiability- A function f(x) is said to be differentiable at x=a if

exists where ‘a’ belongs to I

Rolle’s theorem-

Suppose f(x) is a function defined on [a , b] and it satisfies the following conditions

1. f(x) is continuous in [a , b]

2. f(x) is differentiable in (a , b)

3. f(a) = f(b)

Then there exists at least a point point c ϵ (a , b) , where a<b , such that f’(c) = 0

Proof: suppose y = f(x) is a function and A(a , f(a)) , B(b , f(b)) be two points on the curve f(x) and a,b are two end points. Now conditions for Rolle’s theorem-

1.f(x) is a continuous function in[a , b] , from the figure without breaks in between A&B on y = f(x).

2. f(x) is differentiable in (a , b), because joining A and B we get a line AB.

Slope of the line AB=0 then a point C at P also a tangent at P, or Q,R,S is parallel to x –axis.

Slope of the tangent at P or Q,R,S , will be 0 ,even the curve y = f(x) decreases or increases, that means f(x) is constant.

Derivative of f(x),

f’(c) = 0

That’s why, f’(c) = 0

3. The slope of the line AB is equal to zero, that means the line AB is parallel to x-axis.

So that, f(a) = f(b)

Example: Verify Rolle’s theorem for the function f(x) = x(x+3) in interval [-3 , 0].

Sol. First we will differentiate the given function with respect to x, we get

f’(x) = (x²+3x) + (2x + 3)

This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.

Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).

Here f(x) satisfies all the conditions of Rolle’s theorem,

Then,

f’(x) = 0 , which gives

= 0

We get,

X = 3 and x = -2

Here we can see that clearly -3<-2<0 , therefore there exists -2 ∈ (-3,0) such that

f’(-2) = 0

That means the Rolle’s theorem is true for the given function.

Example: Verify Rolle’s theorem for the given functions below-

1. f(x) = x³ - 6x²+11x-6 in the interval [1,3]

2. f(x) = x²-4x+8 in the interval [1,3]

Sol. (1)

As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = f(3) = 0

Now we find f’(x) = 0

3x² - 12x +11 = 0

We get, x = 2+ and 2 -

Hence both of them lie in (1,3).

Hence the theorem holds good for the given function in interval [1,3]

(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5

Hence f(1) = f(3)

Now the first derivative of the function,

f’(x) = 0

2x – 4 = 0 , gives

X = 2

We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.

This means that the Rolle’s theorem holds good for the given function and given interval.

Example: Verify the Rolle’s theorem for sin x in the interval []

Sol. Suppose f(x) = sin x

We know that sin x is continuous for all x.

Now , f’(x) = cos x exists for all x in () and

f(0

Thus f(x) satisfies all the conditions of Rolle’s theorem.

Now,

f’(x) = 0 that gives , cos x = 0

x =

Here we notice that both intervals lie in (.

There exists, c =

So that, f’(c) = 0

The Rolle’s theorem has been verified.

Lagrange’s mean value theorem-

Suppose that f(x) be a function of x such that,

1. If it is continuous in [a , b]

2. If it is differentiable in (a , b)

Then there atleast exists a value cϵ (a , b)

f’(c) =

Proof:

Lets define a function g(x),

g(x) = f(x) – Ax ………………..(1)

Here A is a constant which is to be determined,

So that, g(a) = g(b)

Now,

g(a) = f(a) – Aa

g(b) = f(b) – Ab

So,

g(a) = g(b),

f(a) – Aa = f(b) – Ab ,

Which gives,

A = …………………..(2)

As right hand side of eq.(1) is continuous in [a,b] , so that g(x) is continuous.

And right hand side of eq.(1) is differential in (a,b) , so that g(x) is differentiable in (a,b).

And g(a) = g(b) , because of the choice of A.

Hence g(x) satisfies all the conditions of Rolle’s theorem.

So that,

There exists a value c such that a<c<b at which g’(c) = 0

Now, differentiate eq. (1) with respect to x, we get

g’(x) = f’(x) – A

Here we know that, x = c,

g’(c) = f’(c) – A

As g’(c) = 0, then

f’(c) – A =0

So that,, f’(c) = A,

From equation (2) , we get

f’(c) = hence proved.

Example: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].

Sol. As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).

f(x) = (x-1)(x-2)(x-3)

f(x) = x-6x²+11x-6

Now at x = 0, we get

f(0) = -6 and

At x = 4, we get.

f(4) = 6

Diff. The function w.r.t.x , we get

f’(x) = 3x²-6x+11

Suppose x = c, we get

f’(c) = 3c²-6c+11

By Lagrange’s mean value theorem,

f’(c) = = = = 3

Now we get,

3c²-6c+11 = 3

3c²-6c+8 = 0

On solving the quadratic equation, we get

C = 2

Here we see that the value of c lies between 0 and 4

Therefore the given function is verified.

Example: : Verify Lagrange’s mean value theorem for f(x) = log xin [1,e].

Sol. We already know that the function which is log x is continuous for all x>0.

So that this is the continuous function In [1,e]

Now,

f’(x) = 1/x

Which is exists for all x in (1,e)

So that f(x) is differentiable in (1,e).

By Lagrange’s mean value theorem, we get

f’(c) = , let x = c,

Then ,

f’(c) =

We get,

c = e-1

e-1 will always lies between 1 and e .

Hence the function is verified by Lagrange’s mean value theorem.

4.2. Taylor’s and Maclaurin’s theorem with reminders

Taylor’s theorem for a function f(x) can be expressed as an infinite series as follows-

f(x) = f(a) + (x – a)f’(a) + + ……….

Example: Expand the function in ascending powers of (x – 1) by using Taylor’s theorem.

Sol. We know that the Taylor’s theorem for the function f(x) in ascending powers of (x – a) is,

f(x) = f(a) + (x – a)f’(a) + + ………. ………….(1)

Here f(x) = and a = 1

Now,

f’(x) = and f’(a) = e

f’’(x) = and f’’(a) = e

Put these values in equation (1),

…………….

Take ‘e’ as common,

…………….}

Which is the required expansion.

Example: Find the Taylor’s series for the function f(x) = 3x² - 6x + 5 in ascending powers of x-3.

Sol. We know that the Taylor’s theorem for the function f(x) in ascending powers of (x – a) is,

f(x) = f(a) + (x – a)f’(a) + + ………. ………….(1)

Here ,

f(x) = 3x² - 6x + 5 and a = 3. So that f(a) = 14

Now , we will find out the derivatives ,

f’(x) = 6x – 6 and f’(3) = 12

f’’(x) = 6 and f’’(3) = 6

f’’’(x) = 0 and f’’’(x) = 14

Put these values in Taylor’s theorem, we get

3x² - 6x + 5 = 14 + 12(x-3) + 6 +…………. Ans.

4.3. Indeterminate forms & L’Hospital’s rule

Let we have two functions f(x) and g(x) and-

Then-

is an expression of the form , in that case we can say that f(x)/g(x) is an indeterminate for of the type at x = a.

Now, Let we have two functions f(x) and g(x) and-

Then-

is an expression of the form , in that case we can say that f(x)/g(x) is an indeterminate for of the type at x = a.

Some other indeterminate forms are

L’Hospital’s rule for form-

Working steps-

1. Check that the limits f(x)/g(x) is an indeterminate form of type .

(note- we can not apply L’Hospital rule if it is not in indeterminate form)

2. Differentiate f and g separately.

3. Find the limits of the derivatives .if the limit is finite , then it is equal to the limit of f(x)/g(x).

Example-1: Evaluate

Sol. Here we notice that it is an indeterminate form of .

So that , we can apply L’Hospital rule-

Example-2: Evaluate .

Sol. Let f(x) = and g(x) = .

Here we see that this is the indeterminate form of 0/0 at x = 0.

Now by using L’Hospital rule, we get-

= = 1

Note- Suppose we get an indeterminate form even after finding first derivative, then in that case , we use the other form of L’Hospital’s rule.

If we have f(x) and g(x) are two functions such that

If exist or (∞ , -∞), then

Example-3: Evaluate

Sol. Let f(x) = , then

And

= 0

But if we use L’Hospital rule again, then we get-

Example-4: Evaluate

Sol. We can see that this is an indeterminate form of type 0/0.

Apply L’Hospital’s rule, we get

But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-

We get

L’Hospital’s rule for form-

Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that

If has a finite limit, or if it is , then

Theorem- If we have f(x) and g(x) are two functions such that .

If exist or (∞ , -∞), then

Example-5: Find , n>0.

Sol. Let f(x) = log x and g(x) =

These two functions satisfied the theorem that we have discussed above-

So that,

Example-6: Evaluate

Sol. Apply L’Hospital rule as we can see that this is the form of

Example-7:

Note- In some cases like above example, we can not apply L’Hospital’s rule.

Other types of indeterminate forms-

Example-8: Evaluate

Sol. Here we find that-

So that this limit is the form of 0.

Now,

Change to obtain the limit-

Now this is the form of 0/0,

Apply L’Hospital’s rule-

4.4. Maxima and minima

If f(x) is a single valued function defined in a region R then

Maxima is a maximum point if and only if

Minima is a minimum point if and only if

Image result for graph of maxima and minima

Maxima and Minima of a function of two independent variables

Let be a defined function of two independent variables.

Then the point is said to be a maximum point of if

Or =

For all positive and negative values of h and k.

Similarly the point is said to be a minimum point of if

Or =

For all positive and negative values of h and k.

Saddle point:Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.

A point is a saddle point of a function of two variables if