Given the following function $$f(x,y)=2x^3y^4$$ I gotta evaluate $D_{\vec{v}}f(1,-1)$, where $\vec{v}=(-3,4)$.
I know $$\nabla f(x,y)=(6x^2y^4,8x^3y^3)$$ and so $$\nabla f(1,-1)=(6,-8)$$ Also, the versor of $\vec{v}$ is $$\vec{u}=\left(\frac{-3}{5},\frac{4}{5}\right)$$ Then $$D_{\vec{v}}f(1,-1)=(6,-8)\cdot\left(\frac{-3}{5},\frac{4}{5}\right)=-10$$ I want to know if what I did is right because WolframAlpha says the final value is $2$.