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$g(x,y)=f(x^3+3y+2), f:R \rightarrow R$, directional derivative in the direction of the vector $(1,0)$ at the point $(0,0)$

My attempt:

I'm not sure how to prove this is zero? Using the definition I got to $\lim_{h \rightarrow 0} \frac{f(h^3+2)-f(2)}{h}$ but I dont know how to compute that. I'm aware that such a directional derivative is just the first partial derivative of $g$ but I'm not sure how that helps me.

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  • $\begingroup$ Hint: By definition of derivative, $f(h^3+2)\approx f(2)+f'(2)h^3$ for $h$ close to $0$. $\endgroup$
    – Arthur
    Dec 21 '17 at 14:49
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Let's use the definition of directional derivative

$$\frac{\partial g}{\partial \vec v}=\nabla\cdot\vec v=v_1\frac{\partial g}{\partial x}+v_2\frac{\partial g}{\partial y}$$

In your case since $\vec v=(1,0)$:

$$\frac{\partial g}{\partial \vec v}=\nabla\cdot\vec v=1\cdot\frac{\partial g}{\partial x}+0\cdot\frac{\partial g}{\partial y}=\frac{\partial g}{\partial x}=3x^2\cdot f'(x^3+3y+2)$$

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Consider $h(x,y)=x^3+3y+2$, $dh=3xdx+3dy$ implies that $d(f\circ h)_{(x,y)}(1,0)=f'(3x^2+2y+2)(3x)$ and if $(x,y)=(0,0)$ the result is $0$.

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