Question

If y = sin^-1(6x√(1 - 9x^2)), - 1/3√2 < x < 1/3√2, then find dy/dx.

Answer 1

Concept:

$\frac{d(sin^-1x)}{dx}=\frac{1}{\sqrt{1-x^2}}$

I have applied substitution method

to find the derivative of given function.

$y = sin^{-1}(6x\sqrt{1-9x^2}})\\\\y = sin^{-1}(2(3x)\sqrt{1-(3x)^2}})$

Take,

$3x=sin\theta\\\\\theta=sin^{-1}(3x)\\\\y = sin^{-1}(2sin\theta\sqrt{1-sin^2\theta}})\\\\y = sin^{-1}(2sin\theta\sqrt{cos^2\theta}})\\\\y = sin^{-1}(2sin\theta.cos\theta})\\\\y = sin^{-1}(sin2\theta})\\\\y = 2\theta\\\\y = 2sin^{-1}(3x)$

Differentiate with respect to 'x'

$\frac{dy}{dx}=2.\frac{1}{\sqrt{1-(3x)^2}}.3\\\\\frac{dy}{dx}=\frac{6}{\sqrt{1-9x^2}}$

Answer 2

Ans:

This is the answer:

Also note that x lies within the domain of sine bijection.