Question

If $x=\sqrt{a^{{\sin }^{-1_t}}}$ and $y=\sqrt{a^{{\cos }^{-1_t}}}$ Find $\frac{dy}{dx}$.

Answer 1

$x=a^{{\sin }^{-1}t},y=a^{{\cos }^{-1}t}$

To find $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{dy}{dt}.\frac{dt}{dx}$

$y=\sqrt{a^{{\cos }^{-1}t}};\frac{dy}{dt}=\frac{d\left(\sqrt{a^{{\cos }^{-1}t}}\right)}{dt}$

$\frac{dy}{dt}=\frac{1}{2\sqrt{a^{{\cos }^{-1}t}}}.\frac{d\left(a^{{\cos }^{-1}t}\right)}{dt}[\frac{d\left(a^x\right)}{d\theta }=a^x\log a]$

$\frac{dy}{dt}=\frac{1}{2\sqrt{a^{{\cos }^{-1}t}}}.a^{{\cos }^{-1}t}.\log a\times -\frac{1}{\sqrt{1-t^2}}$

$\frac{dy}{dt}=-\frac{a^{{\cos }^{-1}t}}{2\sqrt{a^{{\cos }^{-1}t}}}.\frac{\log a}{\sqrt{1-t^2}}....\left(1\right)$

$x=\sqrt{a^{{\sin }^{-1}t}};\frac{dx}{dt}=\frac{d\left(\sqrt{a^{{\sin }^{-1}t}}\right)}{dt}$

$\frac{dx}{dt}=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}.\frac{d\left(a^{{\sin }^{-1}t}\right)}{dt}[\frac{d\left(a^x\right)}{d\theta }=a^x\log a]$

$\frac{dx}{dt}=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}.a^{{\sin }^{-1}t}.\log a\times -\frac{1}{\sqrt{1-t^2}}$

$\frac{dy}{dt}=\frac{a^{{\sin }^{-1}t}}{2\sqrt{a^{{\sin }^{-1}t}}}.\frac{\log a}{\sqrt{1-t^2}}....\left(2\right)$

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=-\frac{a^{{\cos }^{-1}t}}{2\sqrt{a^{{\cos }^{-1}t}}}.\frac{\log a}{\sqrt{1-t^2}}\times \frac{a^{{\sin }^{-1}t}}{2\sqrt{a^{{\sin }^{-1}t}}}.\frac{\log a}{\sqrt{1-t^2}}\Rightarrow \frac{dy}{dx}=\frac{-y}{x}$

Hence proved