PROVE THAT PROBLEMS IN DERIVATIVES

Question 1 :

Differentiate the following

If y = sin-1 x/√(1-x2) show that (1-x2)y2 - 3xy1 - y = 0

Solution :

 y = sin-1 x/√(1-x2)

√(1-x2) y  =  sin-1 x

√(1-x2) y' + y(1/2√(1-x2) )(-2x)  =  1/√(1-x2

√(1-x2) y' + y(-x/√(1-x2) )  =  1/√(1-x2

((1-x2) y' - xy)/√(1-x2)  =  1/√(1-x2

(1-x2) y' - xy  =  1

(1-x2) y'' + y'(-2x) - [xy' + y(1)]  =  0

(1-x2) y'' - 2xy' - xy' - y  =  0

(1-x2) y'' - 3xy' - y  =  0

(1-x2) y2 - 3xy1 - y  =  0

Hence proved.

Question 2 :

If x = a (θ + sin  θ), y = a (1 - cos  θ) then prove that at  θ = π/2, y''  =  1/a

Solution :

x = a (θ + sin  θ)

dx/dθ  =  a (1 + cos θ)--(1)

y = a (1 - cos  θ)

dy/dθ  =  a (0 + sin θ)

=  a  sin θ---(2)

dy/dx  =  (a  sin θ) / a(1 + cos θ)

y'  =  dy/dx  =   sin θ / (1 + cos θ)

y'  =   2sin (θ/2) cos (θ/2) / (2cos2 (θ/2))

y'  =   tan (θ/2) 

d2y/dx2  =  sec2(θ/2) (1/2) (dθ/dx)

By applying (1), we get

d2y/dx2  =  sec2(θ/2) (1/2) (1/(a (1 + cos θ)))

  =  sec2(π/4) (1/2) /(a (1 + cos π/2)))

  =  ((√2)2/2) /(a (1 + 0))

  =  1/a

Hence proved.

Question 3 :

If sin y = x sin(a + y), then prove that dy/dx  =  sin2 (a + y)/sin a , a ≠ nπ

Solution :

 sin y = x sin(a + y)

  x  =  sin y / sin(a + y)

Differentiate with respect to "y"

dx/dy =  [sin (a + y) cos y - sin y cos (x + y)] / sin2(a + y)

dx/dy =  sin (a + y - y)/ sin2(a + y)

dx/dy =  sin a / sin2(a + y)

dy/dx  =  sin2(a + y) / sin a

Hence proved.

Question 4 :

If y = (cos-1x)2 prove that (1 -x2)(d2y/dx2) - x (dy/dx)  - 2 = 0

Solution :

 y = (cos-1x)2

dy/dx  =  2cos-1x (-1/√(1-x2)

dy/dx  =  -2cos-1x /√(1-x2)

In order to remove square root, we may take squares on both sides.

(dy/dx)2  =  (-2cos-1x /√(1-x2))2

(dy/dx)2  =  4(cos-1x)2 /(1-x2)

(dy/dx)2  =  4y2/(1-x2)

(1-x2)(dy/dx)2  =  4y

(1-x2) 2 (dy/dx) (d2y/dx2) + (dy/dx)2 (-2x)  =  4(dy/dx)

Divide by 2 on both sides

(1-x2)(dy/dx) (d2y/dx2) - x(dy/dx)  =  2(dy/dx)

Divide by dy/dx

(1-x2)(d2y/dx2) - x(dy/dx)   =  2

(1-x2)(d2y/dx2) - x(dy/dx) - 2  =  0

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