Question

Prove the formula given in Table 6 for the derivative of each of the following functions.

(a) \({\bf{sec}}{{\bf{h}}^{ - {\bf{1}}}}\) (b) \({\bf{csc}}{{\bf{h}}^{ - {\bf{1}}}}\)

Short answer

(a) \(\frac{1}{{x\sqrt {1 - {x^2}} }}\)

(b) \( - \frac{1}{{\left| x \right|\sqrt {{x^2} + 1} }}\)

Step-by-step solution

Step 1:Find answer for part (a)

Let \(y = {{\mathop{\rm sech}\nolimits} ^{ - 1}}x\), then

\({\mathop{\rm sech}\nolimits} y = x\)

Differentiate the equation \({\mathop{\rm sech}\nolimits} y = x\) with respect to x.

\(\begin{aligned} - {\mathop{\rm sech}\nolimits} y\tanh y\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) &= 1\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= - \frac{1}{{{\mathop{\rm sech}\nolimits} y\tanh y}}\\ &= \frac{1}{{{\mathop{\rm sech}\nolimits} y\sqrt {1 - {{{\mathop{\rm sech}\nolimits} }^2}y} }}\\ &= \frac{1}{{x\sqrt {1 - {x^2}} }}\end{aligned}\)

Therefore, \({{\mathop{\rm sech}\nolimits} ^{ - 1}}x = \frac{1}{{x\sqrt {1 - {x^2}} }}\).

Find answer for part (b)

Let \(y = {{\mathop{\rm csch}\nolimits} ^{ - 1}}x\), then \({\mathop{\rm csch}\nolimits} y = x\).

Differentiate the equation \({\mathop{\rm csch}\nolimits} y = x\) with respect to x.

\(\begin{aligned} - {\mathop{\rm csch}\nolimits} y\coth y\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) &= 1\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= - \frac{1}{{{\mathop{\rm csch}\nolimits} y\coth y}}\\ &= - \frac{1}{{{\mathop{\rm csch}\nolimits} y\left( { \pm \sqrt {{{{\mathop{\rm csch}\nolimits} }^2}y + 1} } \right)}}\\ &= - \frac{1}{{\left| x \right|\sqrt {{x^2} + 1} }}\end{aligned}\)

Therefore, \({{\mathop{\rm csch}\nolimits} ^{ - 1}}x = - \frac{1}{{\left| x \right|\sqrt {{x^2} + 1} }}\).