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Chain Rule of Differentiation

Differentiate...

Question

Differentiate $y = [\frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}]$ w.r.t $x$

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Solution

$\frac{d y}{d x} = \frac{(e^{2 x} - e^{- 2 x}) \frac{d}{d x} (e^{2 x} + e^{- 2 x}) - (e^{2 x} + e^{- 2 x}) \frac{d}{d x} (e^{2 x} - e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= \frac{(e^{2 x} - e^{- 2 x}) (2 e^{2 x} - 2 e^{- 2 x}) - (e^{2 x} + e^{- 2 x}) (2 e^{2 x} + 2 e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= \frac{2 (e^{2 x} - e^{- 2 x}) (e^{2 x} - e^{- 2 x}) - 2 (e^{2 x} + e^{- 2 x}) (e^{2 x} + e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= 2 \frac{{(e^{2 x} - e^{- 2 x})}^{2} - {(e^{2 x} + e^{- 2 x})}^{2}}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= 2 \frac{(e^{2 x} - e^{- 2 x} + e^{2 x} + e^{- 2 x}) (e^{2 x} - e^{- 2 x} - e^{2 x} - e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= 2 \frac{(e^{2 x} + e^{2 x}) (- e^{- 2 x} - e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= 2 \frac{(2 e^{2 x}) (- 2 e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= - 8 \frac{e^{2 x - 2 x}}{{(e^{2 x} - e^{- 2 x})}^{2}}$
$= \frac{- 8}{{(e^{2 x} - e^{- 2 x})}^{2}}$

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