Find dy/dx given x = arccos 1/(sqrt(1+t²)), y = arcsin

Find $\dfrac{dy}{dx}$ given $x = \arccos\dfrac{1}{\sqrt{1+t^{2}}}$, $y = \arcsin\dfrac{t}{\sqrt{1+t^{2}}}$.

Step 1: Differentiate $x$

With $u = \bigl(1+t^{2}\bigr)^{-1/2}$ we have $u' = -t\bigl(1+t^{2}\bigr)^{-3/2}$ and

\[ 1-u^{2} = 1 - \frac{1}{1+t^{2}} = \frac{t^{2}}{1+t^{2}} \;\Longrightarrow\; \sqrt{1-u^{2}} = \frac{|t|}{\sqrt{1+t^{2}}}. \]

The absolute value is the crux. Using $\bigl(\arccos u\bigr)' = -\dfrac{u'}{\sqrt{1-u^{2}}}$, the two minus signs cancel: \[x'_{t} = \frac{t\bigl(1+t^{2}\bigr)^{-3/2}}{|t|\bigl(1+t^{2}\bigr)^{-1/2}} = \frac{t}{|t|\bigl(1+t^{2}\bigr)} = \frac{\operatorname{sgn}(t)}{1+t^{2}}.\]

Step 2: Differentiate $y$

With $v = t\bigl(1+t^{2}\bigr)^{-1/2}$, the product rule gives $v' = \bigl(1+t^{2}\bigr)^{-3/2}$, and

\[ 1-v^{2} = 1-\frac{t^{2}}{1+t^{2}} = \frac{1}{1+t^{2}} \;\Longrightarrow\; \sqrt{1-v^{2}} = \frac{1}{\sqrt{1+t^{2}}}. \]

Here no absolute value appears, since $1+t^{2} \gt 0$: \[y'_{t} = \frac{v'}{\sqrt{1-v^{2}}} = \bigl(1+t^{2}\bigr)^{-3/2}\cdot\sqrt{1+t^{2}} = \frac{1}{1+t^{2}}.\]

Step 3: Form the ratio

The common factor $\dfrac{1}{1+t^{2}}$ cancels, leaving only the sign: \[\frac{dy}{dx} = \frac{y'_{t}}{x'_{t}} = \frac{1}{\operatorname{sgn}(t)} = \operatorname{sgn}(t).\]

Step 4: What the curve actually is

Setting $\theta = \arctan t$, we have $\cos\theta = \dfrac{1}{\sqrt{1+t^{2}}}$ and $\sin\theta = \dfrac{t}{\sqrt{1+t^{2}}}$. Hence $y = \theta = \arctan t$ always, while

\[ x = \arccos\frac{1}{\sqrt{1+t^{2}}} = |\theta| = \bigl|\arctan t\bigr|, \]

because the arccosine returns a value in $[0,\pi]$. So the curve consists of the two rays $y = x$ (for $t \gt 0$) and $y = -x$ (for $t \lt 0$), both emanating from the origin.

Step 5: Verification

\[t = 1:\quad x = \arccos\tfrac{1}{\sqrt2} = \tfrac{\pi}{4},\quad y = \arcsin\tfrac{1}{\sqrt2} = \tfrac{\pi}{4} \;\Longrightarrow\; y = x \;\;\checkmark\]

\[t = -1:\quad x = \tfrac{\pi}{4},\quad y = -\tfrac{\pi}{4} \;\Longrightarrow\; y = -x \;\;\checkmark\]