Differentiate f ( x ) with respect to g ( x ) where f x = sec - 1 1 2 x 2 - 1 , g x = 1 - x 2 .

Given, f x = sec - 1 1 2 x 2 - 1 , g x = 1 - x 2 Put x = cos θ ⇒ θ = cos - 1 x ⇒ f = sec - 1 1 2 cos 2 θ - 1 = sec - 1 ( sec 2 θ ) = 2 θ = 2 cos - 1 x ⇒ d f d x = 2 · - 1 1 - x 2 Now, g = 1 - x 2 ⇒ d g d x = d d x 1 - x 2 = 1 2 1 - x 2 · d d x 1 - x 2 = 1 2 1 - x 2 - 2 x ⇒ d g d x = 1 1 - x 2 - x ∴ d f d g = d f d x d g d x = 2 x

If 1 - x 2 + 1 - y 2 = a x - y , then prove that d y d x = 1 - y 2 1 - x 2 .

Given, 1 - x 2 + 1 - y 2 = a x - y Put x = sin θ , y = sin φ ⇒ θ = sin - 1 x , φ = sin - 1 y ∴ 1 - sin 2 θ + 1 - sin 2 φ = a sin θ - sin φ ⇒ cos θ + cos φ = a ( sin θ - sin φ ) ⇒ 2 cos θ + φ 2 · cos θ - φ 2 = a 2 cos θ + φ 2 · sin θ - φ 2 ⇒ cos θ - φ 2 = a · sin θ - φ 2 ⇒ tan θ - φ 2 = 1 a ⇒ θ - φ 2 = tan - 1 1 a ⇒ φ = θ - 2 tan - 1 1 a ⇒ sin - 1 y = sin - 1 x - 2 tan - 1 1 a Differentiating w.r.t x , we get 1 1 - y 2 · d y d x = 1 1 - x 2 ∴ d y d x = 1 - y 2 1 - x 2 Hence, proved.

If y = x a 2 + x 2 + a 2 log x + a 2 + x 2 , then prove that d y d x = 2 a 2 + x 2 .

Given, y = x a 2 + x 2 + a 2 log x + a 2 + x 2 Differentiating w.r.t x , we get d y d x = d d x x a 2 + x 2 + a 2 log x + a 2 + x 2 ⇒ d y d x = x · 1 2 a 2 + x 2 2 x + a 2 + x 2 · 1 + a 2 x + a 2 + x 2 1 + 1 2 a 2 + x 2 2 x = x 2 a 2 + x 2 + a 2 + x 2 + a 2 x + a 2 + x 2 · a 2 + x 2 + x a 2 + x 2 = x 2 a 2 + x 2 + a 2 + x 2 + a 2 a 2 + x 2 = x 2 + a 2 a 2 + x 2 + a 2 + x 2 = a 2 + x 2 + a 2 + x 2 ∴ d y d x = 2 a 2 + x 2 Hence, proved.

If x log y = log x , then prove that d y d x = y x 1 - log x log y log 2 x .

Given, x log y = log x Applying log on both sides, ⇒ log x log y = loglog x ⇒ ( log y ) ( log x ) = log ( log x ) Differentiating w.r.t x , we get d d x log y log x = d d x log log x ⇒ log x · 1 y · d y d x + log y · 1 x = 1 log x · 1 x ⇒ log x y · d y d x = 1 x log x - 1 x · log y ⇒ log x y · d y d x = 1 - log x · log y x log x ∴ d y d x = y x 1 - log x · log y ( log x ) 2 Hence, proved.