Find the directional derivative at $(1,0,0)$ of the function $$f(x,y,z) = x^2 + ye^z)$$ in the direction of the tangent vector at $g(0)$ to the curve $\mathbb{R}^3$ defined parametrically by $$g(t) = (3t^2 + t + 1, 2t , t^2)$$
$$\begin{align} \nabla f(x,y,z) &= (2x, e^z, ye^z) \\ \frac{\partial f}{\partial g}(1,0,0) &= (2x, e^z, ye^z) \cdot (et^2 + t + 1, 2t, t^2) \\ &= 6t^2 + 2t + 2 + 2t \\ &= 6t^2 + 4t + 2 \end{align}$$
My textbook says that the answer is $2\sqrt{5}$, which I don't think makes any sense and I think the book is wrong, but would like some other input on the matter, thank you
We have
$$g'(t) = (6t + 1, 2 , 2t) \implies g'(0)=(1,2,0)\implies \vec t=\frac1{\sqrt 5}(1,2,0)$$
then
$$\frac{\partial f}{\partial g}(1,0,0) =\nabla f(1,0,0)\cdot \vec t= \frac1{\sqrt 5}(2,1,0)\cdot(1,2,0)=\frac4{\sqrt 5}=\frac45 \sqrt 5 $$