Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 432: 44

$${\text{False}}$$

$$\eqalign{ & {\text{Counterexample}} \cr & {\text{Let }}y = \sqrt {{x^2} + 1} ,{\text{ this is not a rational function}} \cr & {\text{Differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {{x^2} + 1} } \right] \cr & \frac{{dy}}{{dx}} = \frac{{2x}}{{2\sqrt {{x^2} + 1} }} \cr & \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} + 1} }},{\text{ this is a rational function}} \cr & {\text{Therefore,}} \cr & {\text{The statement is false}} \cr} $$