UNIT -1:

Q1:  Find  the nth  derivative of  

Solution:

Since  

Differentiating both side with respect to x

[

Again differentiating with respect to x

Again differentiating with respect to x

   Similarly the nth derivative is

Q2:  Find the  nth  derivative of

Solution:

    Let 

]

Differentiating with  respect  to   x  we get

Again differentiating with respect to x we get

Again differentiating with respect to x we get

Similarly Again differentiating  with  respect to x  we get

Q3: Find the nth  derivative

Solution:

  Let

   Differentiating with respect to x.

Again differentiating with respect to x.

Again differentiating with respect to x.

Again differentiating with respect to x.

Again differentiating with respect to x.

  Similarly the  nth derivative with respect to  x.

Q4: Expand the polynomial in power of , by Taylor’s theorem.

Solution:

Let .

Also

Then

Differentiating with respect  to x.

Again differentiating with respect to x the above function.

Again differentiating with respect to x the above function.

Also the value of above functions at x=2 will be

By Taylor’s theorem

On substituting above values we get

Q5: Expand in power of

Solution:

Let

Also  

Differentiating f(x) with respect to x.

Again differentiating f(x) with respect to x.

Again differentiating f(x) with respect to x.

Also the value of above functions at x=1 will be

By Taylor’s theorem

On substituting above values we get

=

Q6: Expand in power of. Hence find the value of correct to  four decimal places.

Solution:

Let

And .

Differentiating with respect to x.

Again differentiating with respect to x.

Again differentiating with respect to x.

Again differentiating with respect to x.

Also the value of above functions at will be

By Taylor’s theorem

On substituting above values we get

At

.

Q7: Ifusing Taylor’s theorem, show that for .

Solution:

Deduce that

Let then

Differentiating with respect to x.

.Then

Again differentiating with respect to x.

  Then  

Again differentiating with respect to x.

  Then  

By Maclaurin’s theorem

Substituting the above values we get

Since

Hence   

Q8:  Prove that

Solution:

Let 

Differentiating   with respect to x.

Again differentiating with respect to x.

Again  differentiating with respect to x.

Again differentiating with  respect to x.

  and so on.

Putting , in above derivatives we get

so on.

By Maclaurin’s theorem

+………

Substituting the above values we get

Q9:Prove  that

Solution:

Let

Differentiating above function with respect to x.

Again differentiating above function with  respect to x.

Again differentiating above function with  respect to x.

Again differentiating above function with  respect to x.

Putting , in above derivatives we get

so on.

By Maclaurin’s theorem

+………

Substituting the above values we get

Q10:

Solution:

= = 0

Q11:

=?

Solution:

=

=

=

=-2

Q 12:

Find the radius of curvature for the following

Solution:

2y’’-4=0

Y’= =

y’’= = -

y’’.y2 = -2

y’’=-

Substitute y=4, consider y’ and y’’ to solve for R:

We know that,

Radius of convergence (R) = …..(1)

Now substituting the values y’ and y’’ in (1)

=   = 22.36

Radius of convergence (R) =22.36