Test: Differentiation of Implicit Function (19 August) - JEE MCQ

(cosx − sinx)/(cosx + sinx) = (1 − tanx)/(1 + tanx)
tan(A − B) = (tanA − tanB)/(1 + tanAtanB)
= tan(π/4−x)
putting this value in question.
tan⁻¹ tan(π/4−x)
π/4 − x.

so d(π/4 − x)/dx = -1

y = sec⁻¹ x
by rewriting in terms of secant,
⇒ secy = x
by differentiating with respect to x,
⇒ secy tany y'= 1 by dividing by secy tany,
⇒ y'= 1/secy tany
since secy = x and tanx = √sec²y − 1 = √x²−1
⇒ y'= 1/x√x²−1
Hence, d/dx(sec⁻¹x) = 1/x√x²−1

y + sin y = 5x
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)

Identity

xy = 2
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x

3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5 
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
    so let, 3/5 =   cosA
             ⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
     ⇒ (cosAsinxy + sinAcosxy) = 1
     ⇒ sin(A+xy) = 1
     ⇒ A + xy = 2πk + π/2 (k is any integer)
     ⇒ sin⁻¹(4/5) + xy = 2πk + π/2
     differenciating both sides with respect to x
   0 + xdy/dx + y = 0
      dy/dx = -y/x

y = tan^-1(1-cosx)/sinx
y = tan^-1{2sin²(x/2)/(2sin(x/2)cos(x/2)}
y = tan^-1{tan x/2}
y = x/2  => dy/dx = 1/2

cos^-1[(1-x²)/(1+x²)]
Put x = tanθ
y = cos^-1(1 - tan²θ)/(1 + tan²θ)
y = cos^-1(cos²θ)
y = 2θ……………….(1)
Put θ = tan^-1 x in eq(1)
y = 2(tan^-1 x)
dy/dx = 2(1/(1+x²))
dy/dx = 2/(1+x²)

now differentiate y with respect to x,
dy/dx = -{[x d/dx(1+x) - (1+x)dx/dx]}/(1+x²)
= -1/(1+x)²