$\displaystyle Directional\ derivative\ of\ d\ is\ given\ by, \\[2ex] \displaystyle \nabla{}d=i\frac{\partial{}d}{\partial{}x}+j\frac{\partial{}d}{\partial{}y}+k\frac{\partial{}d}{\partial{}z} \\[2ex] \displaystyle \therefore{}\nabla{}d=i\frac{\partial{}\left(x^2y\cos z\right)}{\partial{}x}+j\frac{\partial{}\left(x^2y\cos z\right)}{\partial{}y}+k\frac{\partial{}\left(x^2y\cos z\right)}{\partial{}z} \\[2ex] \displaystyle \therefore{}\nabla{}d=i2xy\cos z+jx^2\cos{z+k}x^2y\left(-\sin z\right) \\[2ex] $

$ Now\ at\ point\ (1,\ 2,\frac{\pi{}}{2}) \\[2ex] \displaystyle \nabla{}d=i(2\bullet{}1\bullet{}2\bullet{}0)+j(1\bullet{}0)+k(1\bullet{}2\bullet{}-1) \\[2ex] \displaystyle \therefore{}\nabla{}d=i\left(0\right)+j\left(0\right)+k(-2) \\[2ex] \displaystyle \therefore{}\nabla{}d=0i+0j-2k \\[2ex] $

$\displaystyle Now\ t=2i+3j+2k \\[2ex] \displaystyle \therefore{}\bar{T}=\frac{2i+3j+2k}{\sqrt{2^2+3^2{+2}^2}} \\[2ex] \displaystyle \therefore{}\bar{T}=\frac{2i+3j+2k}{\sqrt{17}} \\[2ex] $

$\displaystyle Now\ directional\ derivative\ d\ in\ the\ direction\\[2ex] t=2i+3j+2k\ is, \\[2ex] $

$\displaystyle \nabla{}d\bullet{}\bar{T}=\left(0i+0j-2k\right)\bullet{}\left(\frac{2i+3j+2k}{\sqrt{17}}\right) \\[2ex] \displaystyle \therefore{}\nabla{}d\bullet{}\bar{T}=(2)\bullet{}\left(\frac{-2}{\sqrt{17}}\right) \\[2ex] \displaystyle =\ \frac{-4}{\sqrt{17}} \\[2ex] $