Assuming the derivatives of $sin h x$ and $cos h x$ , use the quotient rule to prove that if $y = tan h x = \frac{sin h x}{cos h x}$ then $\frac{d y}{d x} = sec h^{2} x$ .

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Solution

$tan h x = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

now,

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

$\frac{4}{(e^{x} + e^{- x})^{2}} = sec h^{2} x$

hence,

$\frac{d (tan h x)}{d x} = sec h^{2} x$

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