Question

Find the derivative. Simplify where possible.

53. \(y = x{\sinh ^{ - 1}}\left( {\frac{x}{3}} \right) - \sqrt {9 + {x^2}} \)

Short answer

The derivative of function is \(y' = {\sinh ^{ - 1}}\left( {\frac{x}{3}} \right)\).

Step-by-step solution

Write the formula of inverse hyperbolic functions

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

Find the derivative of the function

Consider the function\(y = x{\sinh ^{ - 1}}\left( {\frac{x}{3}} \right) - \sqrt {9 + {x^2}} \).

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

\(\begin{aligned}\frac{d}{{dx}}\left( y \right) & = \frac{d}{{dx}}\left( {x{{\sinh }^{ - 1}}\left( {\frac{x}{3}} \right) - \sqrt {9 + {x^2}} } \right)\\ & = \frac{d}{{dx}}\left( {x{{\sinh }^{ - 1}}\left( {\frac{x}{3}} \right)} \right) - \frac{d}{{dx}}\left( {\sqrt {9 + {x^2}} } \right)\\ & = {\sinh ^{ - 1}}\left( {\frac{x}{3}} \right)\frac{d}{{dx}}\left( x \right) + x\frac{d}{{dx}}\left( {{{\sinh }^{ - 1}}\left( {\frac{x}{3}} \right)} \right) - \frac{1}{{2\sqrt {9 + {x^2}} }}\frac{d}{{dx}}\left( {9 + {x^2}} \right)\\ & = {\sinh ^{ - 1}}\left( {\frac{x}{3}} \right) \cdot 1 + x\frac{{\frac{1}{3}}}{{\sqrt {1 + {{\left( {\frac{x}{3}} \right)}^2}} }} - \frac{1}{{2\sqrt {9 + {x^2}} }}\left( {2x} \right)\\ & = {\sinh ^{ - 1}}\left( {\frac{x}{3}} \right) + \frac{x}{{\sqrt {9 + {x^2}} }} - \frac{x}{{\sqrt {9 + {x^2}} }}\\ & = {\sinh ^{ - 1}}\left( {\frac{x}{3}} \right)\end{aligned}\)

Thus, \(y' = {\sinh ^{ - 1}}\left( {\frac{x}{3}} \right)\).