Question

63-78 Find the derivative of the function. Simplify where possible.

72. \(y = {\bf{ta}}{{\bf{n}}^{ - {\bf{1}}}}\left( {x - \sqrt {{\bf{1}} + {x^{\bf{2}}}} } \right)\)

Short answer

The derivative of the function \(y\) is \(\frac{1}{{2\left( {1 + {x^2}} \right)}}\).

Step-by-step solution

Write the differentiation formula for y

The differentiationof \({\tan ^{ - 1}}\left( x \right)\) is:

\(\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}}\)

Find the derivative of the y

The differentiation of \(y\) is:

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

Simplify further,

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

Thus, the derivative of the function \(y\) is \(\frac{1}{{2\left( {1 + {x^2}} \right)}}\).