Continuity and Differentiability Test - 10

(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 = √sec² y −1 = √x² −1
⇒y '= 1/x √x² −1
Hence, d/dx(sec⁻¹ x) = 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 = 2 πk + π/2 (k is any integer)
     ⇒sin ⁻¹(4/5) + xy = 2 πk + 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 = tan θ
y = cos^-1 (1 - tan² θ)/(1 + tan² θ)
y = cos^-1 (cos² θ)
y = 2 θ……………….(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