Question

To find the derivative of the function.

Short answer

The derivative of the function \(y = x{\tanh ^{ - 1}}x + \ln \sqrt {1 - {x^2}} \) is, \(\frac{{dy}}{{dx}} = {\tanh ^{ - 1}}x\).

Step-by-step solution

 Given function

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

The Concept of derivative and hyperbolic function

Derivative rules:

(1) The chain rule:\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

(2) The derivative of the inverse hyperbolic tangent function is, \(\frac{d}{{dx}}\left( {tan{h^{ - 1}}x} \right) = \frac{1}{{1 - {x^2}}}\).

Evaluate the derivative by the use of concepts

The derivative of the function \(y = x{\tanh ^{ - 1}}x + \ln \sqrt {1 - {x^2}} \) is computed as shown below.

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

Therefore, \(\frac{d}{{dx}}\left( {\tan {h^{ - 1}}x} \right) = \frac{1}{{1 - {x^2}}}\).

Let \(u = \sqrt {1 - {x^2}} \) and apply the chain rule.

\(\begin{aligned}{c}\frac{{dy}}{{dx}} &= \left( {x\left( {\frac{1}{{1 - {x^2}}}} \right) + {{\tanh }^{ - 1}}x} \right) - \frac{d}{{dx}}(\ln u)\\ &= \left( {x\left( {\frac{1}{{1 - {x^2}}}} \right) + {{\tanh }^{ - 1}}x} \right) - \frac{d}{{duu}}(\ln u)\frac{{du}}{{dx}}\\ &= \left( {x\left( {\frac{1}{{1 - {x^2}}}} \right) + {{\tanh }^{ - 1}}x} \right) - \frac{1}{u}\frac{{du}}{{dx}}\end{aligned}\)

Substitute, \(u = \sqrt {1 - {x^2}} \).

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

Therefore, the derivative of the function \(y = x{\tanh ^{ - 1}}x + \ln \sqrt {1 - {x^2}} \) is, \(\frac{{dy}}{{dx}} = {\tanh ^{ - 1}}x.\)