UNIT 2

Differential Calculus-I

2.1. Successive differentiation

Successive differentiation-

The successive differential coefficients of y are denoted as follows-

……………….

The differential coefficient is-

Other notations to denote n’th differential coefficients-

The process of applying differentiation again and again is called successive differentiation.

nth derivative of standard functions-

1. nth derivative of –

Suppose y = ,

Differentiate with respect to x successively, we get

.

.

.

For n times differentiation, we get-

So we can say that its n’th derivative will be

2. nth derivative of log(ax + b)-

Suppose y = log (ax + b)

Differentiate with respect to x successively, we get

(-2)

.

.

.

For n times differentiation, we get-

(-2)…………….(-n + 1)

=  …………….(n - 1)

= (n - 1)

So we can say that its n’th derivative will be

3. nth derivative of -

Suppose y =

Differentiate with respect to x successively, we get

.

.

.

For n times differentiation, we get-

So we can say that its n’th derivative will be

4.  nth derivative of sin(ax + b)-

Suppose y = sin(ax + b)

Differentiate with respect to x successively, we get

.

.

.

For n times differentiation, we get-

So we can say that its n’th derivative will be

Similarly we can find the ‘n’ derivatives of such functions-

Example-1: Find the derivative of the following function-

Sol.  Partial fraction of the function y after splitting-

Suppose x – 1 = z, then

=

=

=

Here we can find its n’th derivative-

Example-2:  If y = , then show that-

Sol. We have,

y =

Differentiate y with respect to x, we get

=

Again diff. (n – 1) times w.r .t x , we get-

2.2. Order differential equation

N’th order differential equations can be solved as below-

Example-1: Find the derivative of

Sol. Here we have-

Suppose,   y =

First derivative-

Here ,

Let x =

So that

Which is the n’th derivative of the given function.

Example-2: Find derivative of the given function:

y =

Sol.   We are given-

y =

Factorize the denominator-

y =

=

derivative will be-

Which is the derivative of the given function.

2.3. Leibnitz’s theorem

If u and v are the functions of x, then-

+……….++…….+ u.

Example-1 If   y   , then prove that-

Sol. Here it is given that-

On differentiating-

Or

= ny.2x

Differentiate again with respect to x, we get-

Or

   …………………. (1)

Differentiate each term of (1) by using Leibnitz’s theorem, we get-

Therefore we get-

Hence proved.

Example-2: If , then prove that-

Sol. Here we have-

Or

Or

y = b cos[ n log(x/n)]              

On differentiating, we get-

Which becomes-

Differentiate again both sides with respect to x, we get-

It becomes-

       ……………….. (1)

Differentiate each term n times with respect to x, we get-

Which is-

         hence proved,

Example-3: If y = , then prove that-

Sol. It is given that-    y =

First derivative –

It becomes-

= 

=

=

Becomes-

+ - = 0

Om differentiating again we get-

+ = 0

Or

+ = 0

Differentiate n times by using Leibnitz’s theorem, we get-

So that

Hence proved.

2.4. Partial differentiation

First order partial differentiation-

Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:

=

Now the partial derivative of f with respect to f can be written as and defined as follows:

=

Note: a. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant.

b. We apply all differentiation rules.

Higher order partial differentiation-

Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.

These are second order four partial derivatives:

(a)     =            

(b)  =                 

(c)   =                 

(d)     =

b and c are known as mixed partial derivatives.

Similarly we can find the other higher order derivatives.

Example-1: - Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

Sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 9x² - 0 + 2y – 8 + 0 – 0

= 9x² + 2y – 8

Similarly partial derivative of f(x,y) with respect to y is:

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 0 – 10y + 2x – 0 + 4 – 0

= 2x – 10y +4.

Example-2: Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

Sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= (y² + 50) cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= 2xy cos(y²x + 5x – 8)

Example-3: Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy⁵)

Sol.    3x²y² - y⁵,      2x³y – 5xy⁴,

  = = 6xy²

  = 2x³ - 20xy³

= = 6x²y – 5y⁴    

= = 6x²y -   5y⁴

Example-4: Find

Sol.  First we will differentiate partially with repsect to r,

Now differentiate partially with respect to θ, we get

Example-5:  if,

Then find.

Sol-

Example-6: if , then show that-

Sol. Here we have,

u =     …………………..(1)

Now partially differentiate eq.(1) w.r to x and y , we get

=

Or

                   ………………..(2)

And now,

=

         ………………….(3)

Adding eq. (1) and (3) , we get

  =  0

Hence proved.

2.5. Partial derivatives of first order and higher orders

(a) Partial Derivative of first order

Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:

=

Now the partial derivative of f with respect to f can be written as and defined as follows:

=

Note: a. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant.

b. We apply all differentiation rules.

(b) Partial Derivative of higher order

Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.

These are second order four partial derivatives:

(a)     =            

(b)  =                 

(c)   =                 

(d)     =

b and c are known as mixed partial derivatives.

Similarly we can find the other higher order derivatives.

Now we will understand the partial derivative by some examples:

Example1- Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

Solution:   To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 9x² - 0 + 2y – 8 + 0 – 0

= 9x² + 2y – 8

Similarly partial derivative of f(x,y) with respect to y is:

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 0 – 10y + 2x – 0 + 4 – 0

= 2x – 10y +4.

Example2 - Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

Solution-   :   To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= (y² + 50) cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= 2xy cos(y²x + 5x – 8)

Example3-   Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy⁵)

Solution-    3x²y² - y⁵,      2x³y – 5xy⁴,

  = = 6xy²

  = 2x³ - 20xy³

= = 6x²y – 5y⁴    

= = 6x²y -   5y⁴

2.6. Homogeneous functions

A function f(x , y) in which the power of each term is the same, is called homogeneous function

The homogeneous function of order n is-

The polynomial of the above function can be written as-

For example- the function-

Is a homogeneous function of order 3.

2.7. Euler’s theorem on homogeneous functions & Relation between second order derivatives of homogeneous functions

Euler’s theorem-

Statement – if u = f(x, y) be a homogeneous function in x and y of degree n , then

x + y = nu

Proof:   Here u is a homogeneous function of degree n,

u = xⁿ f(y/x)        ----------------(1)

Partially differentiate equation (1) with respect to x,

= n f(y/x) + xⁿ f’(y/x).()

Now multiplying by x on both sides, we get

x = n f(y/x) + xⁿ f’(y/x).()  ---------- (2)

Again partially differentiate equation (1) with respect to y,

= xⁿ f’(y/x).

Now multiplying by y on both sides,

y = xⁿ f’(y/x).      ---------------(3)

By adding equation (2) and (3),

xy = n f(y/x) + + xⁿ f’(y/x).()  + xⁿ f’(y/x).

xy         =  n f(y/x)

Here u = f( x, y) is homogeneous function, then -  u = f(y/x)

Put the value of u in equation (4),

xy   =  nu

Which is the Euler’s theorem.

Lets understand Eulers’s theorem by some examples:

Example1-  If u = x²(y-x) + y²(x-y), then show that     -2 (x – y)²

Solution -  here,  u = x²(y-x) + y²(x-y)

u = x²y - x³ + xy² - y³,

Now differentiate u partially with respect to x and y respectively,

  =  2xy – 3x² + y²        --------- (1)

= x² + 2xy – 3y²          ---------- (2)

Now adding equation (1) and (2), we get

= -2x² - 2y² + 4xy

= -2 (x² + y² - 2xy)

= -2 (x – y)²   

Example2-  If u = xy + sin(xy), show that    =  .

Solution –            u = xy + sin(xy)

  =   y+ ycos(xy)

=  x+ xcos(xy)

x (- sin(xy).(y)) + cos(xy)

= 1 – xysin(xy) + cos(xy)     -------------- (1)

1 + cos(xy) + y(-sin(xy) x)

= 1 – xy sin(xy) + cos(xy)      -----------------(2)

From equation (1) and (2),

  = 

Example-3: If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,

Sol. Here we have,

u(x,y,z) = log( tan x + tan y + tan z)  ………………..(1)

Diff. Eq.(1) w.r.t. x , partially , we get

   ……………..(2)

Diff. Eq.(1) w.r.t. y , partially , we get

    ………………(3)

Diff. Eq.(1) w.r.t. z , partially , we get

     ……………………(4)

Now multiply eq. 2 , 3 , 4 by sin 2x , sin 2y , sin 2z respectively and adding , in order to get the final result,

We get,

=

So that, 

                                             hence proved.

2.8. Deductions of Euler’s theorem

Suppose z is a homogeneous function of x,y of degree n and z = f(u), then

Proof-  here z is a homogeneous function of degree n, then we know that by Euler’s theorem.

                 ………………….. (1)

Now z = f(u),

Then,

    and                        

Put these values in equation-1, we get

Which can be written as –

Example-1: If then prove that

Sol. Here, given-

Here u is not a homogeneous function but z =

Cos u, then

Now z is a homogeneous function in x, y of degree 1 / 2.

Now by Euler’s theorem-

Hence proved.

Example-2: If then show that-

Sol. Here we have-

u is not homogenous function here but,

Here z is a function of degree 3.

By using Euler’s deduction formula, we get-

So that

Example-3: If z be a homogeneous function of degree n, then show that-

1.

2.

Sol. (1)We know that by Euler’s theorem-

   ………………………... (1)

Now differentiate equation (1) partially with respect to x, we get-

Proved.

(2)  Differentiate (1) partially with respect to y,

2.9. Composite function

Definition-

If u is a function of x , y and x , y are the functions of t , r, then u is called a composite function of t and r.

Suppose u = f(x , y) and x = g(t , r) , y = h(t , r), then the continuous first order partial derivatives are –

Example: If u = u  then show that

Sol. Here it is given that,

u = u  = u(r , s)

Where r =

Which gives-

     and           …………………. (1)

Now, we know that-

=

=

Or

           ……………….. (2)

Similarly-

Or            

And-

Adding the results, we get

2.10. Asymptotes

An asymptote of a curve of function y = f(x) is a line which does not intersect on the graph.

Types of asymptotes-

1. Vertical asymptote-

A line x = a which is straight is a vertical asymptote of the graph of the function y = f(x) if atleast one of the following condition does it follow-

Example: Determine the vertical asymptote of the function-

Sol. Here we have-

We can write the given function as-

We can notice here as x gets closer to 3 from the left, then the value of the function gets smaller, approaching -∞ and x gets closer 3 from the right, then the value of the function gets larger and approaches +∞.

So that x = 3 is a vertical asymptote same as x = -4 is another vertical asymptote.

2. Horizontal asymptote-

A line y = L is called horizontal asymptote to the curve of the function f(x)-

If f(x)→ L as x→∞ or as x→-∞

Example: Find the horizontal asymptote of the function-

Sol. In order to find the horizontal asymptote-

=

Hence the horizontal asymptote is y = 2.

Example: Find the horizontal asymptote of the function-

Sol. Find the limits-

And

Hence the horizontal asymptotes are y = -1 and y = 5.

Figure will be as follows of the asymptotes of the given function-

3. Procedure to find the asymptotes parallel to axes of a polynomial function-

Theorem-

Suppose f(x , y) is a polynomial in x and y.

A straight line y = c is an asymptote of a curve f(x , y) = 0 if an only if y – c is a factor of the co-efficient of the highest power of x in f(x , y).

Example: Determine the asymptotes parallel to axes of the curve:

y² (x² - a²) = x

Sol. The given function can be written as-

y² (x² - a²) – x = 0

Asymptote parallel to x-axis-

Equating the coefficient of the highest power of x to zero, we get y² = 0 which means y = 0. This is an asymptote.

Asymptote parallel to y-axis-

Equating the coefficient of the highest power of x to zero, we get x² - a² = 0

Which gives x = ±a , that means x = +a , x = -a 

So that these are the asymptotes.

4. Oblique asymptote or slant asymptote-

A straight line y = mx + c where m ≠ 0 will be an oblique/slant asymptote to the graph of the function ‘f’ if-

Note- the value of m can be find as-

And the value of c can be find as-

c =

Example: Find the slant asymptotes of 

Sol. We have,

Which can be written as-

The value of ‘m’ will be-

=     = 

Now we can find ‘c’ as –

c = = =

= 0

Hence the slant asymptotes are

Example: Find the slant asymptote of the function f(x) = x + .

Sol. Find the value of m-

Hence the y = x + c,

Now find c-

Here c must not be infinite.

So we can say that f does not have a slant asymptote at ∞.

2.11. Curve tracing (Cartesian & Polar coordinates)

Curve tracing -  An introduction

A picture explains more clearly than words and numbers. A curve which is the image of functional relationship gives us a lot information about the relation. We can get information analyzing the equation itself but the associated curve is often easy and understandable.

Let’s study about how to trace the curves of various equations in different forms like Cartesian, parametric and polar.

Important definitions for curve tracing

Here are some terms that we use in curve tracing:

1. Double point- when a curve passes two times through this point is known as double point.

2. Node- a double point at which two real tangents can be drawn.(tangents should not be coincide)

3. Cusp -  a double point is a cusp when two tangents are coincide on it.

Tracing a curve- Cartesian form

Let us the equation of a curve is f(x,y)=0 , now we will learn few steps to simplify tracing of this curve.

1. The first step to find out the region of the plane. For example no point in the curve x= y² in second and third quadrant as we will always get a positive value on x-axis. That means our curve will lie on first and fourth quadrant only.

2. The second step is to find out If the curve is symmetrical about any line or origin.

Some examples of symmetrical curves are as below-

                                                                                                         Y

            Y

                                                       X                                                        O                                       X   

Symmetric about the x-axis                                                 Symmetric about origin

Steps to determine the symmetry of a curve:

1. If all the powers of x in function f(x,y)=0 are even, then f(x,y)= f(-x,y) and the curve is always symmetrical about y-axis. Similarly we can draw the conclusion about x-axis

2.If f(x,y)= 0 & f(-x,-y)= 0 , then the curve is symmetrical about the origin.

3. If the equation of the curve does not change when we interchange x and y then it is symmetrical about the line y= x

Let’s understand this with the help of following table-

4. The next step to determine the points where curve intersects the axes. If we put y = 0 in f(x,y)=0 and solve the equation, we get the points interecting on x-axis. Smilarly we get point on y- axis.

5. Now we try to locate the points for discontinuity of the function.

6. Calculate dy/dx to locate the portion where the curve is rising( dy/dx>0) or falling(dy/dx<0)

7. Calculate d²y/dx² to locate maxima and minima and the point of inflection

For maxima = dy/dx = 0,  d²y/dx²<0      &     for minima = dy/dx = 0,  d²y/dx²>0

For point of inflection -  d²y/dx² = 0

8. The next step is to find the asymptotes,

9. Another point to determine the singular point. The shape of the curve at these points generally more complex.

10. Finally plot the points as many as we can. Also try to draw tangents to the curve at some points(calculate the derivative). Now join the plotted point by a smooth curve.

Now we will understand curve tracing with some easy examples:

Example-1:  Trace the curve y = 1/x².

Sol.          As we can see y- coordinates of the curve can not be negative. So the                                              curve must be above x-axis. The curve is also symmetric about y-axis so we can draw the graph only in single side.

Here, we will find the first and second derivatives-

So, dy/dx= -2/x³ and  d²y/dx² = 6/ x⁴    ,  here  dy/dx <0 for all x>0 so we can say that the function is non- increasing so the graph falls as we increase x.

Also second derivative is also non zero so there are no point of inflection.

Here the curve is x²y=1 (rewritten), here both the axes are asymptotes of the curve.

Here is the figure of the curve-:

Example-2: Trace the curve y= (x+2)²(x-1).

Sol. Here first we check the slant asymptotes (there is no vertical asymptote)

So, p =

=

=

=

=  ) = +

Since p is infinity then the function has no slant asymptotes.

Now we will find the point of intersection with axes:

Y(0)= -4;

Put y(x)= 0,     (x+2)²(x-1)= 0

We get by solving,       X1= -2,     x2= 1

Now calculate the first derivative,

Here we will find stationary point by putting first derivative equal to zero:

3x(x+2) = 0

X1 = 0,   x2 = -2

X= -2 is maximum and x= 0 is minimum point.

The value of the function at these points will be:

Y= -4, 0

Now we will find second derivative:

Now, d²y/dx² = 6x+6,

So the function is strictly upward for:

X<-1,

And strictly downward for

x>-1,

Hence x= -1 is the inflection point

We get    y(-1)= -2

Here is the figure of curve:

Tracing a curve- polar form

To trace the polar curve r = f ( we follow the following steps:

1. Symmetry – (a) if the equation is an even function of , then the curve is symmetrical about the initial line.

(b)  if the equation is an even function of r, then the curve is symmetrical about the origin/

(C ). If the equation remains same if we change by - and r by –r then the curve is symmetric about the line through the pole and perpendicular to the initial line.

2. Region- Find the reason for for which r is defined and real

3. Table- make table for the values of r determined by .

4. Angle -   is the angle between the radius vector and tangent to the curve;

= )

Then angle = + , tangents to the curve can be determined by angle

5. Asymptotes-   find the asymptotes of the curve.

Now we will understand of tracing of polar equations by an example:

Example: Trace the curve r = a(1+ cos.

Sol.    Here we can see clearly cos = cos(-), hence the curve is symmetric about initial line.

Since -1, the curve lies inside the circle r = 2a

= -asin     , when  0, thus r decreases as increases in the interval 0 to 

	0
r	2a	a	0

= - cot () = tan(  + ),      this shows the angle between r and and the tangent is 0 or     according to   0 or . Hence the line joining a point on the curve to the origin is orthogonal to the tangent when   = 0 and coincides with

Example-2:  Trace the curve  .

Sol. Symmetry about initial line- Put r = 0,

Hence the straight line are the tangents to the curve at the pole.

Values of ‘r’ as changes from 0 to π-

Table for the values of ‘r’-

	0	0
		Imaginary
	0	0

Here according to the table the curve does not exist for the value lying between  to

The figure will be as follows of the curve-