To find the derivative of \(\sin^{-1}(2x\sqrt{1-x^2})\) with respect to \(\sin^{-1}(3x - 4x^3)\), we need to apply the chain rule of differentiation. The problem can be understood as requiring us to find \(\frac{d}{dx}\left(\sin^{-1}(2x\sqrt{1-x^2})\right)\) and \(\frac{d}{dx}\left(\sin^{-1}(3x - 4x^3)\right)\) separately, and then take their ratio.
- Let \(y = \sin^{-1}(2x\sqrt{1-x^2})\). The derivative of \(\sin^{-1}(u)\) with respect to \(x\) is \(\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx}\).
- Calculate \(\frac{d}{dx}(2x\sqrt{1-x^2})\):
- Let \(u = 2x\sqrt{1-x^2}\).
- Use the product rule: \(u = 2x \cdot \sqrt{1-x^2}\).
- \(\frac{d}{dx}(2x) = 2\).
- \(\frac{d}{dx}(\sqrt{1-x^2}) = \frac{-x}{\sqrt{1-x^2}}\).
- Thus, \(\frac{du}{dx} = 2\sqrt{1-x^2} + (2x)\left(\frac{-x}{\sqrt{1-x^2}}\right) = 2\sqrt{1-x^2} - \frac{2x^2}{\sqrt{1-x^2}}\).
- Simplify to \(\frac{du}{dx} = \frac{2(1-x^2) - 2x^2}{\sqrt{1-x^2}} = \frac{2(1-2x^2)}{\sqrt{1-x^2}}\).
- Substitute back into the derivative formula: \(\frac{d}{dx}\left(\sin^{-1}(2x\sqrt{1-x^2})\right) = \frac{1}{\sqrt{1-(2x\sqrt{1-x^2})^2}} \cdot \frac{2(1-2x^2)}{\sqrt{1-x^2}}.\)
- Simplify:
- Calculate \(1-(2x\sqrt{1-x^2})^2 = 1 - 4x^2(1-x^2) = 1 - 4x^2 + 4x^4\).
- Thus, \(\sqrt{1-u^2} = \sqrt{1 - 4x^2 + 4x^4}\).
- Therefore, the derivative is simplified to \(\frac{2(1-2x^2)}{\sqrt{(1-x^2)(1 - 4x^2 + 4x^4)}}\).
- Now, differentiate \(v = \sin^{-1}(3x - 4x^3)\):
- \(v = 3x - 4x^3\).
- \(\frac{dv}{dx} = 3 - 12x^2\).
- Thus, \(\frac{d}{dx}\left(\sin^{-1}(3x - 4x^3)\right) = \frac{1}{\sqrt{1-(3x - 4x^3)^2}} \cdot (3 - 12x^2).\)
- Find the ratio:
- Let's denote the first derivative as \(dy/dx\) and the second as \(dv/dx\).
- The required derivative \(\frac{dy}{dv} = \frac{dy/dx}{dv/dx}\).
- Substitute the values and simplify to find the ratio. After simplification, it's \(\frac{2}{3}\).
Thus, the derivative of \(\sin^{-1}(2x\sqrt{1-x^2})\) with respect to \(\sin^{-1}(3x - 4x^3)\) is \(\frac{2}{3}\).