I'm trying to find the directional derivative of $h(r,s,t)=ln(3r+6s+9t)$ at (1,1,1) in the direction of $v=4i+2j+6k.$
I have $u=\langle\frac{4}{\sqrt{56}},\frac{2}{\sqrt{56}},\frac{6}{\sqrt{56}}\rangle$ and $F_x=\frac{3}{3r+6s+9t}, F_y=\frac{6}{3r+6s+9t},F_z=\frac{9}{3r+6s+9t}$
then $D_uf=\frac{12}{(3r+6s+9t)\sqrt{56}}+\frac{12}{(3r+6s+9t)\sqrt{56}}+\frac{54}{(3r+6s+9t)\sqrt{56}}=\frac{78}{(3r+6s+9t)\sqrt{56}}$
and at (1,1,1) $=>\frac{78}{18\sqrt{56}}=0.579066...$
The answer is supposed to be $23/42 || 0.5476...$