Question

For $x=\sqrt{a^{{\tan }^{-1}t}},y=\sqrt{a^{{\cot }^{-1}t}},t\in R$, find $\frac{dy}{dx}$.

Answer 1

given

$x=\sqrt{a^{{\tan }^{-1}t}},y=\sqrt{a^{{\cot }^{-1}t}}$

we know that

${\tan }^{-1}t+{\cot }^{-1}t=\frac{\pi }{2}$

therefore subtituting in the equation we get

$y=\sqrt{\frac{a^{\frac{\pi }{2}}}{a^{{\tan }^{-1}x}}}=\sqrt{\frac{a^{\frac{\pi }{2}}}{x^2}}$

$\Rightarrow xy=constant$

$thereforedifferentiatingweget$

$\frac{dy}{dx}=\frac{-y}{x}$