## Calculus 8th Edition

$y=10xln10+10(1-ln10)$
Differentiate $y=10^{x}$ with respect to $x$. $y'=10^{x}ln10$ Let $m$ be the slope of tangent of the line passing through the points (1, 10). $m|_{(1, 10)}=10^{1}ln10 =10 ln10$ The equation of the tangent of the line passing through the points (1, 10) is $(y-10)=10ln10(x-1)$ Hence, $y=10xln10+10(1-ln10)$