Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise: 38

Answer

$y'=\frac{1}{2}(\sqrt x)^{x}(1+lnx)$

Work Step by Step

Given: $y =(\sqrt x)^{x}$ Taking logarithmic on both sides of the function $y =(\sqrt x)^{x}$ $lny=lnx^{\frac{x}{2}}=\frac{x}{2}lnx$ Take implicit differentiation with respect to $x$. Apply product rule of differentiation. $\frac{d}{dx}(lny)=\frac{d}{dx}(\frac{x}{2}lnx)$ $\frac{1}{y}\frac{d}{dx}(y)=\frac{x}{2}\frac{d}{dx}(lnx)+lnx\frac{d}{dx}(\frac{x}{2})$ $\frac{d}{dx}(y)=y[\frac{x}{2}\times(\frac{1}{x})+lnx\times\frac{1}{2}]$ Hence, $y'=\frac{1}{2}(\sqrt x)^{x}(1+lnx)$
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