If x = a t 2 , y = 2 a t , then find d 2 y d x 2 .

Given that we have, x = a t 2 , y = 2 a t Now, differentiate x with respect to t , then we get: d x d t = 2 a t Similarly, differentiate y with respect to t , then: d y d t = 2 a ∴ d y d x = d y d t d x d t = 2 a 2 a t = 1 t Now, differentiate again with respect to x , then: d 2 y d x 2 = - 1 t 2 · d t d x ⇒ d 2 y d x 2 = - 1 t 2 · 1 d x d t ⇒ d 2 y d x 2 = - 1 t 2 · 1 2 a t ∴ d 2 y d x 2 = - 1 2 a t 3 .

If y = e x 2 cos x + ( cos x ) x , then find d y d x .

Given that we have, y = e x 2 cos x + ( cos x ) x Let u = ( cos x ) x ⇒ y = e x 2 · cos x + u Differentiate with respect to x , ∴ d y d x = e x 2 · cos x d d x x 2 · cos x + d u d x ⇒ d y d x = e x 2 · cos x x 2 d d x cos x + cos x d d x x 2 + d u d x ⇒ d y d x = e x 2 · cos x 2 x · cos x - x 2 · sin x + d u d x Now, u = ( cos x ) x ⇒ log u = log ( cos x ) x ⇒ log u = x · log ( cos x ) Now, differentiate with respect to x , then we get: 1 u d u d x = x d d x log ( cos x ) + log cos x d d x x ⇒ 1 u d u d x = x · 1 cos x d d x ( cos x ) + log cos x ⇒ 1 u d u d x = log ( cos x ) - x tan x ⇒ d u d x = ( cos x ) x { log ( cos x ) - x tan x } Hence, d y d x = e x 2 · cos x 2 x · cos x - x 2 · sin x + ( cos x ) x { log ( cos x ) - x tan x } .

If y = e x 2 cos x + ( cos x ) x , then find d y d x .

Given that we have, y = e x 2 cos x + ( cos x ) x Let u = ( cos x ) x ⇒ y = e x 2 · cos x + u Differentiate with respect to x , ∴ d y d x = e x 2 · cos x d d x x 2 · cos x + d u d x ⇒ d y d x = e x 2 · cos x x 2 d d x cos x + cos x d d x x 2 + d u d x ⇒ d y d x = e x 2 · cos x 2 x · cos x - x 2 · sin x + d u d x Now, u = ( cos x ) x ⇒ log u = log ( cos x ) x ⇒ log u = x · log ( cos x ) Now, differentiate with respect to x , then we get: 1 u d u d x = x d d x log ( cos x ) + log cos x d d x x ⇒ 1 u d u d x = x · 1 cos x d d x ( cos x ) + log cos x ⇒ 1 u d u d x = log ( cos x ) - x tan x ⇒ d u d x = ( cos x ) x { log ( cos x ) - x tan x } Hence, d y d x = e x 2 · cos x 2 x · cos x - x 2 · sin x + ( cos x ) x { log ( cos x ) - x tan x } .

If x = a t 2 , y = 2 a t , then find d 2 y d x 2 .

Given that we have, x = a t 2 , y = 2 a t Now, differentiate x with respect to t , then we get: d x d t = 2 a t Similarly, differentiate y with respect to t , then: d y d t = 2 a ∴ d y d x = d y d t d x d t = 2 a 2 a t = 1 t Now, differentiate again with respect to x , then: d 2 y d x 2 = - 1 t 2 · d t d x ⇒ d 2 y d x 2 = - 1 t 2 · 1 d x d t ⇒ d 2 y d x 2 = - 1 t 2 · 1 2 a t ∴ d 2 y d x 2 = - 1 2 a t 3 .

If y = sin ( sin x ) , prove that d 2 y d x 2 + tan x d y d x + y cos 2 x = 0 .

Given that, we have: y = sin ( sin x ) and to prove: d 2 y d x 2 + tan x d y d x + y cos 2 x = 0 Differentiating with respect to x , we get: d y d x = cos ( sin x ) d d x sin x ⇒ d y d x = cos ( sin x ) cos x Again differentiating with respect to x , we get d 2 y d x 2 = - sin ( sin x ) · cos x · cos x - sin x · cos ( sin x ) [Using Product Rule in Differentiation] ⇒ d 2 y d x 2 = - y cos 2 x - sin x d y d x cos x ⇒ d 2 y d x 2 = - y cos 2 x - tan x d y d x ⇒ d 2 y d x 2 + tan x d y d x + y cos 2 x = 0 Hence, proved.

If x = a ( 2 θ - sin 2 θ ) and y = a ( 1 - cos 2 θ ) , find d y d x when θ = π 3 .

Given that we have, x = a ( 2 θ - sin 2 θ ) and y = a ( 1 - cos 2 θ ) Now, differentiate x with respect to θ , then we get: d x d θ = a ( 2 - cos 2 θ · 2 ) ⇒ d x d θ = 2 a ( 1 - cos 2 θ ) Similarly, differentiate y with respect to θ , then d y d θ = a ( sin 2 θ · 2 ) ⇒ d y d θ = 2 a sin 2 θ ∴ d y d x = d y d θ d x d θ = 2 a sin 2 θ 2 a ( 1 - cos 2 θ ) ⇒ d y d x = sin 2 θ 1 - cos 2 θ ⇒ d y d x = 2 sin θ cos θ 2 sin 2 θ ⇒ d y d x = cot θ ∴ d y d x θ = π 3 = cot π 3 = 1 3 .

If x 2 + y 2 2 = x y , find d y d x .

Given that we have, x 2 + y 2 2 = x y Now, differentiate with respect to x , then we get: 2 x 2 + y 2 2 x + 2 y · d y d x = x · d y d x + y [Using Product Rule in Differentiation] ⇒ 4 x x 2 + y 2 + 4 y d y d x x 2 + y 2 = x d y d x + y ⇒ d y d x 4 y x 2 + y 2 - x = y - 4 x x 2 + y 2 ⇒ d y d x = y - 4 x x 2 + y 2 4 y x 2 + y 2 - x ∴ d y d x = y - 4 x 3 - 4 x y 2 4 x 2 y + 4 y 3 - x .

If y = sin - 1 x 2 , prove that 1 - x 2 d 2 y d x 2 - x d y d x - 2 = 0 .

Given y = sin - 1 x 2 Differentiating w.r.t. x , we get ⇒ d y d x = d sin - 1 x 2 d x ⇒ d y d x = 2 sin - 1 x × 1 1 - x 2 ⇒ d y d x = 2 sin - 1 x 1 - x 2 → 1 Again differentiating w.r.t x , we get ⇒ d 2 y d x 2 = 2 d d x sin - 1 x 1 - x 2 = 2 1 - x 2 × d sin - 1 x d x - sin - 1 x × d 1 - x 2 d x 1 - x 2 (Use Quotient Rule in Differentiation) = 2 1 - x 2 × 1 1 - x 2 - sin - 1 x × 1 2 1 - x 2 × ( - 2 x ) 1 - x 2 ⇒ d 2 y d x 2 = 2 1 + x sin - 1 x 1 - x 2 1 - x 2 ⇒ 1 - x 2 d 2 y d x 2 = 2 + 2 x sin - 1 x 1 - x 2 = 2 + x × 2 sin - 1 x 1 - x 2 ⇒ 1 - x 2 d 2 y d x 2 = 2 + x d y d x from 1 ⇒ 1 - x 2 d 2 y d x 2 - x d y d x - 2 = 0 Hence proved.

If x y - y x = a b , find d y d x .

Given x y - y x = a b Let u = x y and v = y x ∴ u - v = a b Differentiate w.r.t. x , we get ⇒ d u d x - d v d x = d d x a b ⇒ d u d x - d v d x = 0 → 1 ∵ d d x ( constant ) = 0 Now, u = x y Taking log on both sides, we get ⇒ log u = y log x Now differentiate w.r.t. x , we get ⇒ 1 u · d u d x = y · 1 x + log x d y d x (Apply product rule of differentiation) ⇒ d u d x = u y x + log x d y d x ⇒ d u d x = x y y x + log x d y d x ⇒ d u d x = x y - 1 · y + x y · log x d y d x Again we have, v = y x Taking log on both sides, we get ⇒ log v = x log y Differentiate w.r.t. x , we get ⇒ 1 v d v d x = x · 1 y d y d x + log y . 1 (Apply product rule of differentiation) ⇒ d v d x = v x y d y d x + log y ⇒ d v d x = y x x y d y d x + log y ⇒ d v d x = x y x - 1 d y d x + y x · log y Now substitute the values of d u d x and d v d x in 1 , we get ⇒ x y - 1 · y + x y · log x d y d x + x y x - 1 d y d x + y x log y = 0 ⇒ x y log x + x y x - 1 d y d x = - x y - 1 · y + y x log y ∴ d y d x = - x y - 1 · y + y x log y x y log x + x y x - 1

If log x 2 + y 2 = 2 tan - 1 y x , show that d y d x = x + y x - y .

Given log x 2 + y 2 = 2 tan - 1 y x Differentiating both sides w.r.t. x , we get d d x log x 2 + y 2 = 2 d d x tan - 1 y x ⇒ 1 x 2 + y 2 × 2 x + 2 y d y d x = 2 × 1 1 + y 2 x 2 × x d y d x - y · 1 x 2 (Use quotient rule and product rule) ⇒ x + y d y d x x 2 + y 2 = x 2 x 2 + y 2 × x d y d x - y x 2 ⇒ x + y d y d x = x d y d x - y ⇒ ( x + y ) = ( x - y ) d y d x ∴ d y d x = x + y x - y Hence proved.