Module 2

Differential Calculus- I

Q 1: Verify Rolle’s theorem for the function f(x) = x2 for

Solution:

Here f(x) = x2;

i)      Since f(x) is algebraic polynomial which is continuous in [-1, 1]

ii)    Consider f(x) = x2

Diff. w.r.t. x we get

f'(x) = 2x

Clearly f’(x) exists in (-1, 1) and does not becomes infinite.

iii)  Clearly

f(-1) = (-1)2 = 1

f(1) = (1)2 = 1

f(-1) = f(1).

Hence by Rolle’s theorem, there exist such that

f’(c) = 0

i.e. 2c = 0

c = 0

Thus such that

f'(c) = 0

Hence Rolle’s Theorem is verified.

Q 2: Verify Rolle’s Theorem for the function f(x) = ex(sin x – cos x) in

Solution:

Here f(x) = ex(sin x – cos x);

i)      Ex is an exponential function continuous for every also sin x and cos x are Trigonometric functions Hence (sin x – cos x) is continuous in and Hence ex(sin x – cos x) is continuous in .

ii)    Consider

f(x) = ex(sin x – cos x)

diff. w.r.t. x we get

f’(x) = ex(cos x + sin x) + ex(sin x + cos x)

    = ex[2sin x]

Clearly f’(x) is exist for each & f’(x) is not infinite.

Hence f(x) is differentiable in .

iii)  Consider

Also,

Thus

Hence all the conditions of Rolle’s theorem are satisfied, so there exist such, that

i.e.

i.e. sin c = 0

But

Hence Rolle’s theorem is verified.

Q 3 Verify whether Rolle’s theorem is applicable or not for

Solution:

Here f(x) = x2;

i)      X2 is an algebraic polynomial hence it is continuous in [2, 3]

ii)    Consider

F’(x) exists for each

iii)  Consider

Thus .

Thus all conditions of Rolle’s theorem are not satisfied Hence Rolle’s theorem is not applicable for f(x) = x2 in [2, 3]

Q 4 Verify the Lagrange’s mean value theorem for

Solution:

Here

i)      Clearly f(x) = log x is logarithmic function. Hence it is continuous in [1, e]

ii)    Consider f(x) = log x.

Diff. w.r.t. x we get,

Clearly f’(x) exists for each value of & is finite.

Hence all conditions of LMVT are satisfied Hence at least

Such that

i.e.

i.e.

i.e.

i.e.

since e = 2.7183

Clearly c = 1.7183

Hence LMVT is verified.

Q5: Verify mean value theorem for f(x) = tan-1x in [0, 1]

Solution:

Here ;

i)      Clearly is an inverse trigonometric function and hence it is continuous in [0, 1]

ii)    Consider

diff. w.r.t. x we get,

Clearly f’(x) is continuous and differentiable in (0, 1) & is finite

Hence all conditions of LMVT are satisfied, Thus there exist

Such that

i.e.

i.e.

i.e.

i.e.

Clearly

Hence LMVT is verified.

Q 6: Prove that

And hence show that

Solution:

Let ; 

i)      Clearly is a logarithmic function and hence it is continuous also

ii)    Consider

diff. w.r.t. x we get,

Clearly f’(x) exist and finite in (a, b) Hence f(x) is continuous and differentiable in (a, b). Hence by LMVT

Such that

i.e.

i.e.

since

a < c < b

i.e.

i.e.

i.e.

i.e.

Hence the result

Now put a = 5, b = 6 we get

Hence the result

Q 7: Prove that , use mean value theorem to prove that,

Hence show that

Solution:

i)      Let f(x) = sin-1x; 

ii)    Clearly f(x) is inverse trigonometric function and hence it is continuous in [a, b]

iii)  Consider f(x) = sin-1x

diff. w.r.t. x we get,

Clearly f’(x) is finite and exists for . Hence by LMVT, such that

i.e.

since a < c < b

i.e.

i.e.

i.e.

i.e.

Hence the result

Put we get

i.e.

i.e.

i.e.

i.e.

Hence the result

Q 8: Use LaGrange’s mean value theorem to determine a point P on the curve y=

where the tangent is parallel to the chord joining (2,0) and (3,1)

solution:

consider y= in [2,3]

(i)                Function is continuous in[2,3] as  algebraic expression with positive exponent is continuous.

(ii)              y’= ,  y’ exists in (2,3) hence the function is derivable in (2,3)

hence the condition of LMV theorem is satisfied.

Hence, there exists one c in (2,3) such that =

  = 4(c-2) = 1 4c=9  c= 4/9

for x = 9/4, tangent is parallel to the chord joining (2,0) and (3,1)

Substituting in (i) we get,

Y=   = = ½

Q 9: Verify LaGrange’s mean value theorem for the following function

f(x) =

putting x=a=2 and x=b=5 ,we get

f(2) =

f(5) =

clearly,

f(2) f(5)

since f(x) is a polynomial function in x, then f(x) is continuous in [2,5].

And f(x) is polynomial in x, then it can be differentiate such that f’(x) = 4x-7

Then by LMV theorem there exists c (2,5) such that’

 f’ (c) =

4c-7 =

c=3.75

Hence Lagrange’s mean value theorem is verified for f(x) in [2,5].

Q 10: Verify Cauchy mean value theorems for &in

Solution:

Let &;

i)      Clearly f(x) and g(x) both are trigonometric functions. Hence continuous in

ii)    Since &

diff. w.r.t. x we get,

&

Clearly both f’(x) and g’(x) exist & finite in . Hence f(x) and g(x) is derivable in and

iii)   

Hence by Cauchy mean value theorem, there exist at least such that

i.e.

i.e. 1 = cot c

i.e.

clearly

Hence Cauchy mean value theorem is verified.

Q 11: Considering the functions ex and e-x, show that c is arithmetic mean of a & b.

Solution:

i)      Clearly f(x) and g(x) are exponential functions Hence they are continuous in [a, b].

ii)    Consider &

diff. w.r.t. x we get

and

Clearly f(x) and g(x) are derivable in (a, b)

By Cauchy’s mean value theorem such that

i.e.

i.e.

i.e.

i.e.

i.e.

i.e.

Thus

i.e. c is arithmetic mean of a & b.

Hence the result

Q 12: Show that

Prove that if

and Hence show that

Verify Cauchy’s mean value theorem for the function x2 and x4 in [a, b] where a, b > 0

If for then prove that,

[Hint:, ]

Q 13: Find the nth derivative of sin3 x

  Sol: 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

 Q 14: Find the nth derivative of sin 5x. sin 3x.?

Sol: let y =   sin 5x.sin 3x= ( sin 5x.sin 3x)  

            ⇒y= (  cos 2x -  cos8x)

           ⇒ y= ( cos 2x- cos8x )

 Differentiate n times w.r.t x,

 Yn  = ( cos 2x -  cos8x )

⇒ yn = ( 2 n (cos( 2x+ nπ/2)-  8n .cos (8x + nπ/2)) nϵz.

Q 15: If  y  = ae n x +  be –nx ,    then show that   y2= n2y

 Sol: Y= aenx + be-nx

  y 1 =   a.n.enx  - b.n.e-nx

 y2 = an2 enx – bn2 e-nx  = n2 (ae nx+ be –nx)

 y2= n2y.

Q 16:  If  y= e-kx/2(a cosnx+ b sinnx) then show that.,y2+ ky1+(n2+ k2/4)y =0

Sol : 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.

Q 17: If  Y 2)  =  log( x + 2) ) then show that (1 + x 2) y1  +xy =1

Sol:  Y 2)  =  log( x + 2) )

Differentiating w.r.to x.,

y.1/ 2  2 .2x+ 2. Y1.

 =(1+x ) y +xy = . 1/x 2  . 2)  +x/ 2 =1

Q 18: Sove the following using lebinitz rule:

Solution:

=0

Q 19: Find the envelope of the family of the ellipses defined by the equation

Solution:

The equation for the given family of curves can be writeen as

=1…..(1)

Where the semi axis a is the parameter and 0<a<1

Differentiating eq(1) w.r.to a  is

Take the square root of both sides of the equation:

Express from the last equation

Substituting in the eq(1) we get

=1

=

Q 20: Find the evolute for the following parabola  y=

Solution:

For evoluting the parabola we have the following formulae,

Substituting the given function we get

By eliminating the x in we getx =

Substituting in we get

Therefore the required answer is

Q 21: For a given curve  convert the point (2,) from polar to Cartesian form

Solution:

  r=2 and =   we have,

x = rcos and y= rsin

x= 2.cos = 2.1/2 = 1

y = 2.sin = 2./2 =

Therefore the required Cartesian point is (1,).