1. **Problem Statement:** Find the differential coefficient of the function $$y = x^{\sin^{-1} x}$$ with respect to $$\sin^{-1} x$$. 2. **Understanding the problem:** We want to find $$\frac{dy}{d(\sin^{-1} x)}$$, the derivative of $$y$$ with respect to $$\sin^{-1} x$$. 3. **Rewrite the function:** Let $$\theta = \sin^{-1} x$$, so $$x = \sin \theta$$. Then the function becomes: $$y = x^{\theta} = (\sin \theta)^{\theta}$$. 4. **Use logarithmic differentiation:** Take natural logarithm on both sides: $$\ln y = \theta \ln (\sin \theta)$$. 5. **Differentiate both sides with respect to $$\theta$$:** $$\frac{1}{y} \frac{dy}{d\theta} = \ln (\sin \theta) + \theta \frac{1}{\sin \theta} \cos \theta$$ 6. **Simplify the derivative:** $$\frac{dy}{d\theta} = y \left( \ln (\sin \theta) + \theta \cot \theta \right)$$ 7. **Substitute back $$y$$ and $$\theta$$:** $$\frac{dy}{d(\sin^{-1} x)} = x^{\sin^{-1} x} \left( \ln x + (\sin^{-1} x) \frac{\sqrt{1 - x^2}}{x} \right)$$ Note: We used $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$ and $$\cos \theta = \sqrt{1 - x^2}$$ since $$x = \sin \theta$$. **Final answer:** $$\boxed{\frac{dy}{d(\sin^{-1} x)} = x^{\sin^{-1} x} \left( \ln x + (\sin^{-1} x) \frac{\sqrt{1 - x^2}}{x} \right)}$$