Show that the transformation z = logtan x 2 reduces the differential equation d 2 y d x 2 + cosec x c o t x d y d x + 4 y cosec 2 x = 0 into d 2 y d z 2 + 4 y = 0 .

Given z = logtan x 2 d z d x = 1 tan x 2 · sec 2 x 2 · 1 2 = 1 2 sin x 2 cos x 2 = 1 sin x = cosec x d y d x = d y d z · d z d x = cosec x d y d z Now, d 2 y d x 2 = d d x d y d x = d d x cosec x d y d z = - cosec x cot x d y d z + cosec x d 2 y d z 2 · d z d x = - cosec x cot x d y d z + cosec 2 x d 2 y d z 2 Given, d 2 y d x 2 + cot x cosec x d y d x + 4 y cosec 2 x = 0 - cosec x cot x d y d z + cosec 2 x d 2 y d z 2 + cot x cosec x d y d x + 4 y cosec 2 x = 0 cosec 2 x d 2 y d z 2 + 4 y = 0 ⇒ d 2 y d z 2 + 4 y = 0

Show that the transformation z = logtan x 2 reduces the differential equation d 2 y d x 2 + cosec x c o t x d y d x + 4 y cosec 2 x = 0 into d 2 y d z 2 + 4 y = 0 .

Given z = logtan x 2 d z d x = 1 tan x 2 · sec 2 x 2 · 1 2 = 1 2 sin x 2 cos x 2 = 1 sin x = cosec x d y d x = d y d z · d z d x = cosec x d y d z Now, d 2 y d x 2 = d d x d y d x = d d x cosec x d y d z = - cosec x cot x d y d z + cosec x d 2 y d z 2 · d z d x = - cosec x cot x d y d z + cosec 2 x d 2 y d z 2 Given, d 2 y d x 2 + cot x cosec x d y d x + 4 y cosec 2 x = 0 - cosec x cot x d y d z + cosec 2 x d 2 y d z 2 + cot x cosec x d y d x + 4 y cosec 2 x = 0 cosec 2 x d 2 y d z 2 + 4 y = 0 ⇒ d 2 y d z 2 + 4 y = 0

Differentiate the following function w.r.t. x : 1 - cos 2 x 1 + cos 2 x

Let y = 1 - cos 2 x 1 + cos 2 x As we know, cos 2 x = 2 cos 2 x - 1 = 1 - 2 sin 2 x ⇒ y = 2 sin 2 x 2 cos 2 x = tan 2 x ⇒ y = tan x Using the property of modulus function, y = - tan x - π 2 < x < 0 tan x 0 ≤ x < π 2 Differentiating above equation with respect to x , d y d x = - sec 2 x - π 2 < x < 0 sec 2 x 0 ≤ x < π 2 Again using the property of modulus function, d y d x = sec 2 x tan x tan x .

Differentiate the following function w.r.t. x : sec - 1 x + 1 x - 1 + sin - 1 x - 1 x + 1 .

Let y = sec - 1 x + 1 x - 1 + sin - 1 x - 1 x + 1 As we know, sec - 1 x = cos - 1 1 x ⇒ y = cos - 1 x - 1 x + 1 + sin - 1 x - 1 x + 1 ⇒ y = π 2 ∵ sin - 1 x + cos - 1 x = π 2 Differentiating above equation with respect to x , d y d x = d d x π 2 = 0 .

Differentiate the following w.r.t. x : 2 x cot x x .

Let y = 2 x cot x x Differentiating above equation with respect to x , d y d x = d d x 2 x cot x x Applying quotient rule of differentiation, d y d x = x d d x 2 x cot x - 2 x cot x d d x x x 2 ∵ d d x f x g x = g x f ' x - f x g ' x g 2 x Applying product rule of differentiation, d y d x = x 2 x d d x cot x + cot x d d x 2 x - 2 x cot x d d x x x 2 ∵ d d x f x g x = f ' x g x + g ' x f x ⇒ d y d x = x 2 x - c o s e c 2 x + cot x 2 x log e 2 - 2 x cot x 1 2 x x = 2 x - c o s e c 2 x x + cot x 2 x log e 2 x - 2 x cot x 1 2 x x ∵ d d x cot x = - c o s e c 2 x , d d x a x = a x ln a , d d x x n = n x n - 1 = 2 x x - cosec 2 x + cot x log e 2 - cot x 2 x .

Differentiate following w.r.t. x : sin x - x cos x x sin x + cos x

Let y = sin x - x cos x x sin x + cos x Differentiating above equation with respect to x , d y d x = d d x sin x - x cos x x sin x + cos x Applying quotient rule of differentiation, d y d x = x sin x + cos x d d x sin x - x cos x - sin x - x cos x d d x x sin x + cos x x sin x + cos x 2 ∵ d d x f x g x = g x f ' x - f x g ' x g 2 x d y d x = x sin x + cos x cos x - d d x x cos x - sin x - x cos x d d x x sin x - sin x x sin x + cos x 2 ∵ d d x sin x = cos x , d d x cos x = - sin x Applying product rule of differentiation we get, d y d x = x sin x + cos x cos x - cos x - x sin x - sin x - x cos x x cos x + sin x - sin x x sin x + cos x 2 ∵ d d x f x g x = f ' x g x + g ' x f x d y d x = x sin x + cos x x sin x - sin x - x cos x x cos x x sin x + cos x 2 d y d x = x 2 sin 2 x + cos x x sin x - x cos x sin x + x 2 cos 2 x x sin x + cos x 2 d y d x = x 2 sin 2 x + x 2 cos 2 x x sin x + cos x 2 d y d x = x 2 x sin x + cos x 2 ∵ sin 2 x + cos 2 x = 1