Question

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

(a) \({\bf{cos}}{{\bf{h}}^{ - {\bf{1}}}}\) (b) \({\bf{tan}}{{\bf{h}}^{ - {\bf{1}}}}\) (c) \({\bf{cot}}{{\bf{h}}^{ - {\bf{1}}}}\)

Short answer

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

(b) \(\frac{1}{{1 - {x^2}}}\)

(c) \(\frac{1}{{1 - {x^2}}}\)

Step-by-step solution

Step 1:Find an answer for part (a)

Let \(y = {\cosh ^{ - 1}}x\), then \(\cosh y = x\).

Differentiate the equation \(\cosh y = x\) with respect to x.

\(\begin{aligned}\sinh y\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) &= 1\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{1}{{\sinh y}}\\ &= \frac{1}{{\sqrt {{{\cosh }^2}y - 1} }}\\ &= \frac{1}{{\sqrt {{x^2} - 1} }}\end{aligned}\)

Therefore, \({\cosh ^{ - 1}}x = \frac{1}{{\sqrt {{x^2} - 1} }}\).

Find an answer for part (b)

Let \(y = {\tanh ^{ - 1}}x\), then \(\tanh y = x\).

Differentiate the equation \(\tanh y = x\) with respect to x.

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

Therefore, \({\tanh ^{ - 1}}x = \frac{1}{{1 - {x^2}}}\).

Find an answer for part (c)

Let \(y = {\coth ^{ - 1}}x\), then \(\coth y = x\).

Differentiate the equation \(\coth y = x\) with respect to x.

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

Therefore, \({\coth ^{ - 1}}x = \frac{1}{{1 - {x^2}}}\).