# Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 490: 35

$\dfrac{t^2+1}{2t^2}$

#### Work Step by Step

Given: $G(t)=\dfrac{d}{dt}\sinh {(\ln t)}$ Now, $G'(t)=\dfrac{d}{dt}\sinh {(\ln t)}$ On applying the chain rule, we have $G'(t)=(\dfrac{d\sinh {(\ln t)}}{d\ln t})(\frac{d\ln t}{dt})$ or, $=\cosh {(\ln t)} \times \dfrac{1}{t}$ or, $=\dfrac{\cosh {(\ln t)}}{t}$ Since, $\cosh x=\dfrac{e^x+e^{-x}}{2}$ $G'(t)=\dfrac{\dfrac{e^{\ln t}+e^{-\ln t}}{2}}{t}=\dfrac{t+\frac{1}{t}}{2t}$ or, $=\dfrac{\frac{t^2+1}{t}}{2t}$ or, $=\dfrac{t^2+1}{2t^2}$

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