Differentiate the function w.r.t. x : tan - 1 x 3 - x 1 - 3 x .

The given function is, y = tan - 1 x 3 - x 1 - 3 x Let x = tan t ⇒ t = tan - 1 x y = tan - 1 tan t 3 - tan 2 t 1 - 3 tan 2 t y = tan - 1 3 tan t - tan 3 t 1 - 3 tan 2 t y = tan - 1 tan 3 t ∵ tan 3 A = 3 tan A - tan 3 A 1 - 3 tan 2 A y = 3 t ∵ tan - 1 tan x = x y = 3 tan - 1 x [ Substituting value from t ] On differentiating both sides with respect to x , we get d d x y = d d x 3 tan - 1 x d y d x = 3 1 + x 2 d d x x [ Use chain rule and also tan - 1 x = 1 1 + x 2 ] d y d x = 3 1 + x · 1 2 x ∵ d d x x = 1 2 x d y d x = 3 2 x 1 + x .

Differentiate the function tan - 1 3 a 2 x - x 3 a a 2 - 3 x 2 w.r.t. x .

The given function is, y = tan - 1 3 a 2 x - x 3 a a 2 - 3 x 2 ⇒ y = tan - 1 3 a 2 x - x 3 a 3 - 3 a x 2 Dividing every term by a 3 , we get y = tan - 1 3 a 2 x a 3 - x 3 a 3 a 3 a 3 - 3 a x 2 a 3 y = tan - 1 3 x a - x a 3 1 - 3 x a 2 Put x a = tan t ⇒ t = tan - 1 x a y = tan - 1 3 tan t - tan 3 t 1 - 3 tan 2 t y = tan - 1 tan 3 t ∵ tan 3 x = 3 tan x - tan 3 x 1 - 3 tan 2 x y = 3 t ∵ tan - 1 tan x = x y = 3 tan - 1 x a [ Substituting value from t ] On differentiating both sides with respect to x , we get d d x y = d d x 3 tan - 1 x a d y d x = 3 · 1 1 + x a 2 d d x x a [ Use chain rule and also d d x tan - 1 x = 1 1 + x 2 ] d y d x = 3 1 + x 2 a 2 · 1 a ∵ d d x x n = n x n - 1 d y d x = 3 a 2 + x 2 a 2 · 1 a d y d x = 3 a a 2 + x 2 .

If y = 1 + sin x 1 + cos x 1 + sin x 1 + cos x 1 + . . . . . to ∞ , prove that d y d x = 1 + y cos x + y sin x 1 + 2 y + cos x - sin x .

The given function is, y = 1 + sin x 1 + cos x 1 + sin x 1 + cos x 1 + . . . . . to ∞ Let y = 1 + sin x 1 + cos x 1 + . . . . . to ∞ ⇒ y = sin x 1 + cos x 1 + y ⇒ sin x 1 + y + cos x 1 + y ⇒ y = sin x 1 + y 1 + y + cos x ⇒ y + y 2 + y cos x = sin x 1 + y Differentiating both side with respect to x , we get d d x y + y 2 + y cos x = d d x sin x 1 + y d d x y + d d x y 2 + d d x y cos x = sin x · d d x 1 + y + 1 + y d d x sin x [ Using product rule ] d y d x + 2 y · d y d x + y · d d x cos x + cos x · d d x y = sin x · d d x 1 + d d x y + 1 + y · cos x [ Using product rule and chain rule and also d d x f x + g x = d d x f x + d d x g x ] d y d x + 2 y · d y d x + y · - sin x + cos x · d y d x = sin x · 0 + d y d x + cos x 1 + y ∵ d d x x n = n x n - 1 ; d d x cos x = - sin x d y d x + 2 y · d y d x - y sin x + cos x · d y d x = sin x · d y d x + cos x 1 + y d y d x + 2 y · d y d x + cos x · d y d x - sin x · d y d x = cos x 1 + y + y sin x d y d x 1 + 2 y + cos x - sin x = cos x 1 + y + y sin x [ Taking common d y d x ] d y d x = 1 + y cos x + y sin x 1 + 2 y + cos x - sin x .

If y = log cos x sin x log sin x cos x - 1 , show that d y d x π 4 = - 8 log 2 e

We need to prove that, y = log cos x sin x log sin x cos x - 1 y = log sin x log cos x log cos x log sin x - 1 ∵ log b a = log a log b y = log sin x log cos x log sin x log cos x y = log sin x 2 log cos x 2 y = log sin x log cos x 2 Differentiating both sides with respect to x , we get d d x y = d d x log sin x log cos x 2 d y d x = 2 log sin x log cos x d d x log sin x log cos x [ Use chain rule and also d d x x n = n x n - 1 ] d y d x = 2 log sin x log cos x log cos x d d x log sin x - log sin x d d x log cos x log cos x 2 [ Using quotient rule ] d y d x = 2 log sin x log cos x log cos x 1 sin x d d x sin x - log sin x 1 cos x d d x cos x log cos x 2 d d x log x = 1 x d y d x = 2 log sin x log cos x log cos x cos x sin x - log sin x - sin x cos x log cos x 2 ∵ d d x sin x = cos x ; d d x cos x = - sin x d y d x = 2 log sin x log cos x log cos x cot x + log sin x tan x log cos x 2 ∵ cos x sin x = cot x ; sin x cos x = tan x At x = π 4 d y d x π 4 = 2 log sin π 4 log cos π 4 log cos π 4 cot π 4 + log sin π 4 tan π 4 log cos π 4 2 d y d x π 4 = 2 log 1 2 log 1 2 log 1 2 · 1 + log 1 2 · 1 log 1 2 2 ∵ cos π 4 = 1 2 = sin π 4 ; cot π 4 = 1 = tan π 4 d y d x π 4 = 2 2 log 1 2 log 1 2 2 d y d x π 4 = 4 log 1 2 d y d x π 4 = 4 log e 2 - 1 d y d x π 4 = 4 - 1 2 log e 2 d y d x π 4 = - 8 log 2 e . ∵ log b a = log a b Hence proved.

Differentiate x x sin - 1 x w.r.t. x .

The given function is, y = x x sin - 1 x On differentiating both sides with respect to x , we get d d x y = d d x x x sin - 1 x d y d x = x x · d d x sin - 1 x + sin - 1 x d d x x x [ Using product rule ] d y d x = x x 1 - x 2 d d x x + sin - 1 x d d x x x [ Use chain rule and also d d x sin - 1 x = 1 1 - x 2 ] d y d x = x x 1 - x 1 2 x + sin - 1 x d d x x x ∵ d d x x = 1 2 x d y d x = x x 2 x - x 2 + sin - 1 x x x x · 1 x + log x · 1 ∵ d d x log x = 1 x d y d x = x x 2 x - x 2 + sin - 1 x · x x 1 + log x d y d x = x x 2 x - x 2 + x x 1 + log x sin - 1 x .

If x = sin t , y = sin 2 t , prove that 1 - x 2 d y d x 2 = 4 1 - y 2 .

Given function is, x = sin t … … … 1 y = sin 2 t … … … 2 then d y d x = d y d t d x d t Let x = sin t On differentiating both side with respect to t , we get d d t x = d d t sin t d x d t = cos t ∵ d d x sin x = cos x Let y = sin 2 t On differentiating both side with respect to t , we get d y d t = d d t sin 2 t d y d t = cos 2 t d d t 2 t [ Use chain rule and also d d x sin x = cos x ] d y d t = cos 2 t · 2 ∵ d d x x n = n x n - 1 d y d t = 2 cos 2 t Now, d y d x = d y d t d x d t d y d x = 2 cos 2 t cos t cos t · d y d x = 2 cos 2 t [ Cross multiply cos t ] On squaring both sides, we get cos t · d y d x 2 = 2 cos 2 t 2 cos 2 t · d y d x 2 = 4 cos 2 2 t 1 - sin 2 t · d y d x 2 = 4 1 - sin 2 2 t ∵ cos 2 x = 1 - sin 2 x 1 - x 2 d y d x 2 = 4 1 - y 2 [ Substituting value from sin t and sin 2 t ] Hence, proved.