## Thomas' Calculus 13th Edition

$\lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt$
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa: $f_x=e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} \\ f_y= e^{x+y} \cos z +arcsin x \\f_z=-e^{x+y} \sin z$ Write the gradient vector equation. $\nabla f = \lt f_x,f_y,f_z \gt$ $\implies \nabla f = \lt e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} , e^{x+y} \cos z +arcsin x , -e^{x+y} \sin z \gt$ Thus, $\nabla f (0,0,\dfrac{\pi}{6}) = \lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt$