If y = e^(sin-1 x) and z = e^(-cos-1 x), prove that (dy)/(dz) = e^(π/2)

If $y = e^{\sin^{-1} x}$ and $z = e^{-\cos^{-1} x}$, prove that $\frac{dy}{dz} = e^{\frac{\pi}{2}}$

Given: $y = e^{\sin^{-1} x}$ and $z = e^{-\cos^{-1} x}$

To show: $\frac{dy}{dz} = e^{\frac{\pi}{2}}$

Answer

$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = e^{\pi/2}$.

1. $y = e^{\sin^{-1}x}$ and $z = e^{-\cos^{-1}x}$ (given).
2. $\frac{dy}{dx} = e^{\sin^{-1}x}\cdot\frac{1}{\sqrt{1-x^2}}$ (chain rule, $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$).
3. $\frac{dz}{dx} = e^{-\cos^{-1}x}\cdot\left(-\frac{d}{dx}\cos^{-1}x\right) = e^{-\cos^{-1}x}\cdot\frac{1}{\sqrt{1-x^2}}$ (chain rule, $\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}$).
4. $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{e^{\sin^{-1}x}}{e^{-\cos^{-1}x}}$ (parametric differentiation; the factors $\frac{1}{\sqrt{1-x^2}}$ cancel).
5. $= e^{\sin^{-1}x+\cos^{-1}x}$ (law of indices).
6. $\sin^{-1}x+\cos^{-1}x = \frac{\pi}{2}$ (standard identity).
7. Hence $\frac{dy}{dz} = e^{\frac{\pi}{2}}$.