If y = sin x ° and z = cos x , then find d y d z .

We have: y = sin ⁡ x ° Now differentiate w.r.t x , we get: ⇒ d y d x = cos ⁡ x ° × π 180 and z = cos ⁡ x Now differentiate w.r.t x , we get: d z d x = − sin ⁡ x Hence, d y d z = d y d x d z d x = π 180 × cos ⁡ x ° − sin ⁡ x = - π 180 × cosec x × cos ⁡ x ∘

If y = sin x ° and z = cos x , then find d y d z .

We have: y = sin ⁡ x ° Now differentiate w.r.t x , we get: ⇒ d y d x = cos ⁡ x ° × π 180 and z = cos ⁡ x Now differentiate w.r.t x , we get: d z d x = − sin ⁡ x Hence, d y d z = d y d x d z d x = π 180 × cos ⁡ x ° − sin ⁡ x = - π 180 × cosec x × cos ⁡ x ∘

If y = e sin - 1 x ; z = e - cos - 1 x , show that the value of dy dz is independent of x .

We have, y = e sin − 1 ⁡ x and z = e − cos − 1 ⁡ x Consider, y = e sin − 1 ⁡ x Taking log on both sides, we get: log ⁡ y = sin − 1 ⁡ x Differentiate w.r.t x , we get: 1 y × d y d x = 1 1 - x 2 ⇒ d y d x = y 1 - x 2 Now consider, z = e − cos − 1 ⁡ x Taking log on both sides, we get log ⁡ z = − cos − 1 ⁡ x Differentiate w.r.t x , we get: 1 z × d z d x = 1 1 − x 2 ⇒ d z d x = z 1 − x 2 Hence, d y d z = d y d x d z d x = y 1 − x 2 z 1 − x 2 = y z = e sin − 1 ⁡ x e − cos − 1 ⁡ x = e sin − 1 ⁡ x × e cos − 1 ⁡ x = - e sin − 1 ⁡ x + cos − 1 ⁡ x = e π 2 ∵ sin − 1 ⁡ x + cos − 1 ⁡ x = π 2

If y = x e x , then show that x d y d x = ( x + 1 ) y

We have: y = x e x . . . 1 ⇒ e x = y x . . . 2 Differentiate 1 both sides w.r.t x , we get d y d x = d x e x d x ⇒ d y d x = x × d e x d x + e x × d x d x ⇒ d y d x = x × e x + e x × 1 ⇒ d y d x = x × e x + e x ⇒ d y d x = e x x + 1 ⇒ d y d x = y x × ( x + 1 ) using 2 ⇒ x × d y d x = ( x + 1 ) × y Hence proved.

If y = sec x + tan x , then show that 2 × d y d x = 1 + y 2 .

We have, y = sec ⁡ x + tan ⁡ x Differentiate both sides w.r.t x , we get: d y d x = sec ⁡ x tan ⁡ x + sec 2 ⁡ x ⇒ d y d x = sec ⁡ x ( sec ⁡ x + tan ⁡ x ) ⇒ d y d x = y sec ⁡ x ∵ y = sec ⁡ x + tan ⁡ x Consider, L . H . S 2 × d y d x = 2 × y sec x = 2 y sec x . . . 1 Consider, R . H . S 1 + y 2 = 1 + sec x + tan x 2 ∵ y = sec x + tan x = 1 + sec 2 x + tan 2 x + 2 sec x × tan x = sec 2 ⁡ x + sec 2 ⁡ x + 2 sec ⁡ x × tan ⁡ x ∵ 1 + tan 2 A = sec 2 A = 2 sec 2 ⁡ x + 2 sec ⁡ x × tan ⁡ x = 2 sec ⁡ x sec x + tan ⁡ x = 2 y sec x . . . 2 From 1 , 2 we get the desired result.

If y = x - 5 x + 5 , then show that 2 x d y d x + y 2 = 1 .

We have: y = x − 5 x + 5 Differentiate w.r.t x , we get: d y d x = d d x x − 5 x + 5 ⇒ d y d x = x + 5 × d x - 5 d x - x - 5 × d x + 5 d x x + 5 2 [Using Quotient Rule] ⇒ d y d x = x + 5 × 1 - x - 5 × 1 x + 5 2 ⇒ d y d x = x + 5 - x + 5 x + 5 2 ⇒ d y d x = 10 x + 5 2 Consider, 2 x d y d x + y 2 = 2 x × 10 ( x + 5 ) 2 + x − 5 x + 5 2 = 20 x ( x + 5 ) 2 + x − 5 2 x + 5 2 = 20 x + x − 5 2 ( x + 5 ) 2 = 20 x + x 2 + 25 - 10 x ( x + 5 ) 2 = 10 x + x 2 + 5 2 ( x + 5 ) 2 = 2 × 5 x + x 2 + 5 2 ( x + 5 ) 2 = ( x + 5 ) 2 ( x + 5 ) 2 = 1 Hence proved.

If y = x + a 2 + x 2 n , then show that d y d x = n y a 2 + x 2

We have: y = x + a 2 + x 2 n Taking log on both sides, we get: ⇒ log ⁡ y = n log ⁡ x + x 2 + a 2 ∵ log m n = n log m Differentiating both sides w.r.t x , we get: d log y d x = d n log ⁡ x + x 2 + a 2 d x ⇒ 1 y × d y d x = n x + x 2 + a 2 × d x + x 2 + a 2 d x ⇒ 1 y × d y d x = n x + x 2 + a 2 × d x d x + d x 2 + a 2 d x ⇒ 1 y × d y d x = n x + x 2 + a 2 × 1 + 2 x 2 x 2 + a 2 ⇒ 1 y × d y d x = n x + x 2 + a 2 × x + x 2 + a 2 x 2 + a 2 ⇒ 1 y × d y d x = n x 2 + a 2 ⇒ d y d x = n y x 2 + a 2 Hence proved.

If y = 1 - x 1 + x then show that 1 - x 2 d y d x + y = 0 .

We have: y = 1 − x 1 + x ⇒ y = ( 1 − x ) ( 1 − x ) ( 1 + x ) ( 1 − x ) ⇒ y = ( 1 − x ) 2 ( 1 + x ) ( 1 − x ) ⇒ y = 1 − x 1 − x 2 − ( 1 ) Differentiating both sides w.r.t x , we get: ⇒ d y d x = 1 − x 2 × d d x ( 1 − x ) − ( 1 − x ) d d x 1 − x 2 1 − x 2 2 [Using Quotient rule] ⇒ d y d x = − 1 − x 2 + ( 1 − x ) × x 1 − x 2 1 − x 2 ⇒ 1 − x 2 d y d x = − 1 + x 2 + x − x 2 1 − x 2 ⇒ 1 − x 2 d y d x = − 1 + x 1 − x 2 ⇒ 1 − x 2 d y d x = − 1 − x 1 − x 2 From 1 , we get: 1 − x 2 d y d x = − y ⇒ 1 − x 2 d y d x + y = 0 Hence proved.

If y = x 1 + x 2 , then show that x 3 × d y d x = y 3 .

We have: y = x 1 + x 2 Differentiate both sides w.r.t x , we get d y d x = d d x ( x ) × x 2 + 1 − x × d d x x 2 + 1 x 2 + 1 ⇒ d y d x = x 2 + 1 − x 2 x 2 + 1 × d d x x 2 + 1 x 2 + 1 ⇒ d y d x = x 2 + 1 − x 2 x 2 + 1 × 2 x x 2 + 1 ⇒ d y d x = x 2 + 1 − x 2 x 2 + 1 x 2 + 1 ⇒ d y d x = 1 x 2 + 1 3 2 ⇒ d y d x = y 3 x 3 ∵ y = x x 2 + 1 ⇒ x 3 × d y d x = y 3 Hence proved.

If x = a sin - 1 ⁡ t , y = a cos - 1 ⁡ t , show that d y d x = - y x

Given that x and y are connected parametrically and here, x = a sin - 1 ⁡ t , y = a cos - 1 ⁡ t d x d t = 1 2 a sin - 1 ⁡ t · d d t a sin - 1 ⁡ t = 1 2 a sin - 1 ⁡ t · a sin - 1 ⁡ t · log ⁡ a 1 1 - t 2 = 1 2 x · x 2 · log ⁡ a 1 1 - t 2 = x log ⁡ a 2 1 - t 2 d y d t = 1 2 a cos - 1 ⁡ t · d d t a cos - 1 ⁡ t = 1 2 a cos - 1 ⁡ t · a cos - 1 ⁡ t · log ⁡ a - 1 1 - t 2 = 1 2 y · y 2 · log ⁡ a 1 1 - t 2 = - y log ⁡ a 2 1 - t 2 d y d x = d y d t d x d t = - y log ⁡ a 2 1 - t 2 x log ⁡ a 2 1 - t 2 = - y x (Hence proved)