Find the directional derivative of: f(x,y) = (x^2)y-y^3, (2,1,3) Can somebody explain to me why we did the step we did? First we found the directional derivative of f(x,y,z) ▽f(x,y) = <2xy, (x^2)-3(y^2), a> ▽f(2,1) = <2(2)(1), (2^2)-3(1^2)), a> = <2(2), 4-3, a> = <4,1, a> and then we... did some sorcery mathematics?! f(x,y) = (x^2)y-y^3 z = f(x, y) F(x,y,z) = f(x,y) - z ▽F(x,y,z) = <2xy, (x^2)-3(y^2), -1> ▽F(2,1,3) = <4,1,-1> A dot product for some reason... <4,1,a><4,1,-1> = 0 4(4)+1(1)-a = 0 16+1 = a 17 = a Therefore, the directional derivative is, <4,1,17>. __________________________________________ My question is: if we are given a function f(x,y) and are told to find the directional derivative at the point (xo,yo,zo) is it safe to say that we can always solve it like we did above? Du▽f(x,y,z) = <fx(x,y), fy(x,y), <fx(x,y), fy(x,y)><fx(x,y), fy(x,y)>>? Can somebody explain to me why the equation above is valid? This isn't homework, I'm just reviewing my notes and couldn't understand what happened here. This is just for my own understanding... and because of that I'm worried I'm posting this in the wrong thread. Thank you for taking the time to read my question.