Calculate directional derivative and find equation of a plane tangent to function plot

Question

I'm given the following problem: Calculate directional derivative of a function $$f(x,y) = x^2 - xy + 3y^2$$ at point $$(1,0)$$ in the direction of a vector $$v = (3,4)$$ and provide the equation of a plane tangent to plot of said function at point $$(1, 0, 0)$$ I'd be very thankful if someone could show me how to approach this problem. I know how to calculate partial derivatives, but I'm not sure how to calculate directional derivative and find plane equation.

Answer 1

For differentiable functions, which is the case here, the directional derivative can be computed as $f'_{(v_1,v_2)}(a,b) = v_1 f'_x(a,b) + v_2 f'_y(a,b)$, i.e. $$ f'_{(3,4)}(1,0) = 3 f'_x(1,0) + 4f'_y(1,0)= 3 \times 2 + 4 \times (-1) =2 $$

Regarding the equation of the plane, it will be of the form $$ z-1 = 2(x-1) + (-1)(y-0)\Leftrightarrow 2x-y-z = 1. $$