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 :

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   =  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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