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First Principle of Differentiation

Given that ...

Question

Given that $tan h y = t$ , show that $y = \frac{1}{2} ln (\frac{1 + t}{1 - t})$ for $- 1 < t < 1$

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Solution

Given that $t a n h y = t$

$\Rightarrow \frac{e^{2 y} - 1}{e^{2 y} + 1} = t$

$\Rightarrow e^{2 y} (1 - t) = 1 + t$

$\Rightarrow 2 y = l n (\frac{1 + t}{1 - t})$

$\Rightarrow y = \frac{1}{2} l n (\frac{1 + t}{1 - t})$

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\int \frac{\frac{1}{2}}{1 + t} \Rightarrow \frac{1}{2} ln (1 + t)

x = log (1 + t^{2}), y = t - {tan}^{- 1} t

\frac{d y}{d x} = \frac{\sqrt{e^{x} - 1}}{2}

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

\frac{d y}{d x} = - \frac{y}{x}

x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})

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

x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})

, then show that

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

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