Unit-3

Calculus

Q1)  Differentiate the function f(x) = by using the first principal method.

A1)

We know that-

Here

Substituting ( for x gives-

Hence-

Q2) Evaluate the

A2)

We can simply find the Solutionution as follows,

Q3) Evaluate

A3)
 

Q4) Differentiate with respect to x.

A4)

Let

Now

Q5) if y = then find dy/dx.

A5)

Suppose z =

Now-

So that-

Q6) If y = log loglog then find dy/dx.

A6)

Suppose y = log u where u = log v and v = log

So that-

Q7) Find the derivative of

A7)

Let y = then-

Q8) If y = log x/ x a then find

A8)

First we will find the first derivative-

Now

Q9) if y = then find .

A9)

Here

y =

Then

Q10) Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

A10)

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.

Q11) if , then show that-

A11)

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

Hence proved.

Q12) If u = x²(y-x) + y²(x-y), then show that   -2 (x – y)²

A12)

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

Q13) State and prove Euler’s theorem.

A13)

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.

Q13) If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,

A13)

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, 

Q14) let q = 4x + 3y      and    x = t³ + t² + 1    , y = t³ - t² - t

Then find  .

A14)

. =

Where,

f1 = ,  f2 = 

In this example f1 = 4    ,      f2 = 3

Also,       

3t² + 2t    ,    

4(3t² + 2t) + 3(

=  21t² + 2t – 3

Q15) If u = u( y – z , z - x , x – y)  then prove that  = 0

A15)

Let,

Then,

By adding all these equations we get,

= 0

Hence proved.

Q16) If z is the function of x and y , and x =   , y = , then prove that,

A16)

Here , it is given that,  z is the function of x and y & x , y are the functions of u and v.

So that,

  ……………….(1)

And,

   ………………..(2)

Also there is,

x =   and  y = ,

now,

   ,       ,      ,    

From equation(1) , we get

  ……………….(3)

And from eq. (2) , we get

    …………..(4)

Subtracting eq. (4) from (3), we get

= ) – (

= x

Hence proved.

Q17) Examine for maximum and minimum for the function f(x) =

A17)

Here the first derivative is-

So that, we get-

Now we will get to know that the function is maximum or minimum at these values of x.

For x = 3

Let us assign to x, the values of 3 – h and 3 + h (here h is very small) and put these values at f(x).

Then-

Which is negative for h is very small

Which is positive

Thus f’(x) changes sign from negative to positive as it passes through x = 3.

So that f(x) is minimum at x = 3 and the minimum value is-

And f(x) is maximum at x = -3.

Q18) Find all points of maxima and minima of the function f(x) =.

A18)

Here,

f(x) =.

f’(x) =

Put, f’(x) = 0

We get,

Hence x = -1 and x = 3 are the critical values of the given function.

Now take,

Where x = -1, then f’’(x) = -12

Since f’’(x) < 0 at x = -1, the function has maxima at x = -1

And when x = 3, then f’’(x) = 12

Since f’’(x) > 12 at x = 3, the function has minima at x =

Q19) Find the integral of-
 

A19)

We know that-

Then

Q20) Evaluate the following integral-

A20)

Let us suppose,

Then-

Or

Substituting –

Q21) Evaluate-

A21)

Here according to ILATE,

First function = log x

Second function =

We know that-

Then-

On solving, we get-