Solve \left\{ \begin{array} { l } { x = \sqrt { t ^ {

Question

Rewrite the parametric equations

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

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Find the first derivative

\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

More Steps

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

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