Write the rectangular equation of \left\{ \begin{array} { l } { x = \sqrt{t^{2} + 1} } \\ { y = \ln{\left ( t + \sqrt{t^{2} + 1} \right )} } \end{array} \right.

\begin{align}&y=\ln{\left(\sqrt{x^{2}-1}+\left|x\right|\right)}\\&y=\ln{\left(-\sqrt{x^{2}-1}+\left|x\right|\right)}\end{align}

Question :

begin{array l x = sqrt{t^{2 + 1} } y = ln{ ( t + sqrt{t^{2 + 1} )} } endarray .

y = ln (x^{2} - 1 + ∣ x ∣) y = ln (- x^{2} - 1 + ∣ x ∣)

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\frac{d y}{d x} = \frac{1}{t}

Evaluate

{x = t^{2} + 1 y = ln (t + t^{2} + 1)

To find the derivative \frac{d y}{d x},first find \frac{d x}{d t} and \frac{d y}{d t}

\frac{d}{d t} (x) = \frac{d}{d t} (t^{2} + 1) \frac{d}{d t} (y) = \frac{d}{d t} (ln (t + t^{2} + 1))

Find the derivative

More Steps

\frac{d x}{d t} = \frac{t}{t ^{2} + 1} \frac{d}{d t} (y) = \frac{d}{d t} (ln (t + t^{2} + 1))

Find the derivative

More Steps

\frac{d x}{d t} = \frac{t}{t ^{2} + 1} \frac{d y}{d t} = \frac{1}{t ^{2} + 1}

Find the required derivative by Substituting \frac{d x}{d t} = \frac{t}{t ^{2} + 1} and \frac{d y}{d t} = \frac{1}{t ^{2} + 1} into \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}

\frac{d y}{d x} = \frac{\frac{1}{t ^{2} + 1}}{\frac{t}{t ^{2} + 1}}

Solution

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\frac{d y}{d x} = \frac{1}{t}

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