1.

Differentiate sin-1 (3x - 4x3) w.r.t x

Answer» assuming the given function as

y= sin^-1( 3x- 4x^3)

Let x =sinu

=> dx/du = cosu

then y = sin^-1(3sinu-4sin^3u)=sin^-1(sin 3u) = 3u

dy/du= 3

So dy/dx = (dy/du)/(dx/du)=3/cosu= 3/[(1-sin^2u)^1/2]=3/[(1-x^2)^1/2]

let y = sin-1(3x - 4x3)

let x = sinθ

Then 3x - 4x3 = 3sinθ - 4sin3θ

= sin3θ

Therefore, y = sin-1(3x - 4x3)

=sin-1(sin3θ)

= 3θ

Now, \(\frac{dy}{dx}\) = \(\frac{dy}{dθ}\) . \(\frac{dθ}{dx}\) = 3  \(\frac{dθ}{dx}\) (∵ y = 3θ)

∵ x = sinθ ⇒ \(\frac{dx}{dθ}\) = cosθ (By differentiating w.r.t θ)

⇒\(\frac{dθ}{dx}\) = \(\frac{1}{cosθ}\) = \(\frac{1}{\sqrt{1-sin^2θ}}\) = \(\frac{1}{\sqrt{1-x^2}}\) (∴ siinθ = x)

Therefore \(\frac{dy}{dx}\) = \(3\times\frac{1}{\sqrt{1-x^2}}\) (∴\(\frac{dθ}{dx}\) = \(\frac{1}{\sqrt{1-x^2}}\))

∵ \(\frac{d}{dx}\) sin-1(3x - 4x3) =  \(\frac{3}{\sqrt{1-x^2}}\) (∵ = y = sin-1(3x - 4x3))


Discussion

No Comment Found

Related InterviewSolutions