Question

Find the derivative. Simplify where possible.

52. \(y = x{\tanh ^{ - 1}}x + \ln \sqrt {1 - {x^2}} \)

Short answer

The derivative of the function is \(y' = {\tanh ^{ - 1}}x\).

Step-by-step solution

Write the formula of inverse hyperbolic functions

\(\frac{d}{{dx}}\left( {{{\tanh }^{ - 1}}x} \right) = \frac{1}{{{x^2} - 1}}\)

Find the derivative of the function

Consider the function\(y = x{\tanh ^{ - 1}}x + \ln \sqrt {1 - {x^2}} \).

Differentiate the function w.r.t\(x\).

\(\begin{aligned}\frac{d}{{dx}}\left( y \right) & = \frac{d}{{dx}}\left( {x{{\tanh }^{ - 1}}x + \ln \sqrt {1 - {x^2}} } \right)\\ & = \frac{d}{{dx}}\left( {x{{\tanh }^{ - 1}}x} \right) + \frac{d}{{dx}}\left( {\ln \sqrt {1 - {x^2}} } \right)\\ & = \frac{d}{{dx}}\left( {x{{\tanh }^{ - 1}}x} \right) + \frac{d}{{dx}}\left( {\frac{1}{2}\ln \left( {1 - {x^2}} \right)} \right)\\ & = {\tanh ^{ - 1}}x + \frac{x}{{1 - {x^2}}} + \frac{1}{2}\left( {\frac{1}{{1 - {x^2}}}} \right)\frac{d}{{dx}}\left( {1 - {x^2}} \right)\\ & = {\tanh ^{ - 1}}x + \frac{x}{{1 - {x^2}}} + \frac{1}{2}\left( {\frac{1}{{1 - {x^2}}}} \right)\left( { - 2x} \right)\\{\tanh ^{ - 1}}x\end{aligned}\)

Thus, \(y' = {\tanh ^{ - 1}}x\).