Unit – 1

Function of one variable

Q1) What is Leibnitz’s theorem?

A1)

Successive differentiation-

The successive differential coefficients of y are denoted as follows-

……………….

The differential coefficient is-

Statements of Leibnitz’s Theorem-

If u and v are the function of x  such  that  their nth  derivative exists, then the nth derivative of  their product will  be

Q2) What is nth derivative of ?

A2)

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

Q3) Find the nth derivative of sin(ax + b).

A3)
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

Q4) If y = l , then show that-

A4)

We have,

y =

Differentiate y with respect to x, we get

=

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

Q5) Find cos x cos 2x cos 3x.

A5)

So that-

n’th derivative-

Q6) Find the derivative of the following function-

A6)

Partial fraction of the function y after splitting-

Suppose x – 1 = z, then

=

=

=

Here we can find its n’th derivative-

Q7) Find if

A7)

Here we have-

At x = 0,

When n is odd-

When n is even

Q8) Find the nth derivative of sin3 x.

A8)

We know that sin 3x= 3sin x 4sin3 x =  sin3x  = 

Differentiate   n    times w.r.t x,

( sin3 x) = (3 sinx- sin3x)

=  (  -3n. Sin(  3x+ nπ/2)  +   3 sin (x+  nπ/2)) nϵz

Q9) If  y= e-kx/2(a cosnx+ b sinnx) then show that.,y2+ ky1+(n2+ k2/4)y =0.

A9)

y= e-kx/2(a cosnx+ b sinnx)

Differentiating  w.r.to. x.,

Y1 = e-kx/2( -an sin nx + bn cos nx) - k/2.y

Y1+ k/2.y = ne-kx/2 ( -an sin nx + bn cos nx)    (1)

Differentiating w.r.to x.,

Y2+ k/2.y1 = ne-kx/2  (-k/2) ( -an sin nx + bn cos nx) + n e-kx/2(-an  cosnx- bn  sinnx).

= -(k/2) (y1+ k/2 y)- n2 y = - (k/2 y1)- ( k2/4)y- n2y.

y2 + ky1 +(n2+ k2/4)y = 0.

Q10) If   y  , then prove that-

A10)

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.

Q11) If y = , then prove that-

A11)

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.

Q12) If ,     then show that

A12)

Also, find 

Here

Differentiating with respect to x, we get

   …(ii)

Squaring   both  side  we get

   …(iii)

Again differentiating with  respect to  x ,we get

Using Leibnitz’s  theorem  we  get

        …(iv)

Putting x=0  in equation (i),(ii) and (iii)  we get

Putting  n=1,2,3,4….

………………

Hence   

Q13) What is Maclaurin’s theorem.

A13)

Maclaurin’s series-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)  

We get-

+ ……

Q14) Expand by using Maclaurin’s series.

A14)

Let

Put these values in Maclaurin’s series-

Or

Q15) Expand upto x6

A15)

Here

Now we know that

    … (1)

    … (2)

Adding (1) and (2) we get

Q16) Expand in power of (x – 3)

A16)

Let

Here a = 3

Now by Taylor’s series expansion,

 … (1)

equation (1) becomes.

Q17) Using Taylors series method expand in powers of (x + 2)

A17)

Here

a = -2

By Taylors series,

   … (1)

Since

,  , …..

Thus equation (1) becomes

Q18) Using Taylors series method expand in powers of (x + 2).

A18)

Here

a = -2

By Taylors series,

   … (1)

Since

,  , …..

Thus equation (1) becomes

Q19) Evaluate

A19)

-6.

Q20) Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

A20)

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.

Q21) Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

A21)

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)

Q22) if , then show that-

A22)

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.

Q23) If where then find the value of ?

A23)

Given  

Where 

By chain rule

Now substituting the value of x ,y,z we get

-6

8

Q24) If where the relation is .

Find the value of

A24)

Let the given relation is denoted by

We know that

Differentiating u with respect to x and using chain rule

Q25) State and prove Euler’s theorem.

A25)

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,

= nf(y/x) + xⁿ f’(y/x).()

Now multiplying by x on both sides, we get

x = nf(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 = nf(y/x) + + xⁿ f’(y/x).()  + xⁿ f’(y/x).

xy         =  nf(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.

Q26) If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,

A26)

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.

Q27) Expand f(x , y) = in powers of x and y about origin.

A27)
Here we have the function-

f(x , y) =

Here , a = 0 and b = 0 then

f(0 , 0) =

Now we will find partial derivatives of the function-

Now using Taylor’s theorem-

+………

Suppose h = x and k = y, we get

+…….

= +……….

Q28) Find out the maxima and minima of the function

A28)

Given …(i)

Partially differentiating (i) with respect to x we get

….(ii)

Partially differentiating (i) with respect to y we get

….(iii)

Now, form the equations

Using (ii) and (iii) we get

using above two equations

Squaring both side we get

Or 

This show that

Also we get

Thus we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0)  we get

So, the point (0,0) is a saddle point.

At point we get

So the point is the minimum point where

In case 

So the point is the maximum point where

Q29) If ,Find the value of x and y for  which is maximum.

A29)

Given function is

And relation is

By Lagrange’s Method

[]     ..(i)

Partially differentiating (i) with respect to  x, y and z and equate them tozero

Or       …(ii)

Or       …(iii)

Or       …(iv)

On solving (ii),(iii) and (iv) we  get

Using the given relation  we get

So that

Thus the point for the maximum value of the given function is