Question

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

Short answer

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

Step-by-step solution

Identify function parts

Identify the function parts that could be differentiated individually: \(u = x\), \(v = \tanh ^{-1} x\), and \(w = \ln \sqrt{1-x^{2}}.\)

Differentiate the functions individually

Differentiate the functions individually applying the rules of differentiation. The derivative of \(u = x\) is \(u'=1\). The derivative of \(v = \tanh ^{-1} x\) can be found using the formula for the derivative of the arctanh function which is \(v' = \frac{1}{1-x^2}\). The derivative of \(w = \ln \sqrt{1-x^{2}}\) is found using the chain rule for logarithmic functions, \(w' = \frac{1}{\sqrt{1-x^{2}}} * \frac{-2x}{2\sqrt{1-x^{2}}} = \frac{-x}{1-x^{2}}\).

Apply the product rule

Apply the product rule (uv)' = u'v + uv' to the first part of the function. \(y'_{first} = 1*\tanh ^{-1} x + x*\frac{1}{1-x^2} = \tanh ^{-1} x + \frac{x}{1-x^{2}}\).

Combine the results

Combine the two parts of the derivative found in steps 2 and 3 to get the final derivative of the function. \(y' = y'_{first} + w' = \tanh ^{-1} x + \frac{x}{1-x^{2}} + \frac{-x}{1-x^{2}} = \tanh ^{-1} x\).