## Calculus: Early Transcendentals 8th Edition

y' = -$\frac{1}{x(lnx)^{2}}$
y = $\frac{1}{lnx}$ Recall for this problem: Quotient Rule Differentiate: y' = ($\frac{1}{lnx}$)' y' = $\frac{lnx(1)' - 1(lnx)'}{(lnx)^{2}}$ y' = $\frac{0 - \frac{1}{x}}{(lnx)^{2}}$ y' = $\frac{-\frac{1}{x}}{(lnx)^{2}}$ y' = -$\frac{1}{x(lnx)^{2}}$