Unit - 1

Hyperbolic functions

Q1) What do you understand by the higher order derivatives?

A1)

It is the process of differentiating the given function simultaneously many times and the result obtained are called successive derivative.

Let be a differentiable function.

First derivative is denoted by

Second derivative is  

Third derivative is

Similarly the nth derivative is

Q2) What is the n’th 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 n’th derivative of log(ax + b).

A3)

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

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 the derivative of

A5)

Here we have-

Suppose,  y =

First derivative-

Here ,

Let x =

So that

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

Q6) Find cos x cos 2x cos 3x.

A6)

So that-

n’th derivative-

Q7) Find the derivative of the following function-

A7)

Partial fraction of the function y after splitting-

        Suppose x – 1 = z, then

           =

          =

      =

Here we can find its n’th derivative-

Q8) If   y  , then prove that-

A8)

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-

[ (x2 – 1) yn+2 + nyn+1 (2x)+ n(n-1)/  yn(2)] + 2 (1 – n)[ xyn+1 + nyn] – 2nyn = 0

Therefore we get-

   Hence proved.

Q9) If y = , then prove that-

A9)

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-

yn+2(1 – x2) + nC1 yn+1 (-2x) + nC2 yn(-2) – [yn+1. x+ nC1. yn. 1] + m2yn = 0

(1 – x2)yn+2 – 2nx yn+1 – n(n – 1)yn – xyn+1 – nyn + m2yn = 0

So that

Hence proved.

Q10) If ,     then show that

A10)

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   

Q11) Determine the vertical asymptote of the function-

A11)

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.

Q12) Find the horizontal asymptote of the function-

A12)

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-

Q13) Find the slant asymptotes of 

A13)

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

Q14) Trace the curve y = 1/x².

A14)

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-:

Q15) Trace the curve r = a(1+ cos.

A15)

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

From the above results, we can easily draw the graph above the initial line

Q16) Trace the curve  .

A16)

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-

Q17) Evaluate .

A17)

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

Q18) Evaluate

A18)

Let f(x) = , then

And

= 0

= 0

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

Q19) Evaluate

A19)

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-

Q20) Find the maximum and minimum point of the function

A20)

Partially differentiating given equation with respect to and x and y then equate them to zero                 

       On solving above we get 

         Also

            Thus we get the pair of values (0,0), (,0) and (0,

        Now, we calculate

        At the point (0,0)

       So function has saddle point at (0,0).

       At the point (

         So the function has maxima at this point (.

        At the point (0,

       So the function has minima at this point (0,.

        At the point (

        So the function has an saddle point at (