Derivative of primitive function in high dimension

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Suppose $x \in \mathbb{R}^n$, and $f : \mathbb{R} \rightarrow \mathbb{R}$, $g :\mathbb{R}^n \rightarrow \mathbb{R}$ are two functions. $F(u)$ is the primitive function of $f$, $F(u) = \int_0^u f(v) dv$.

I want to compute the gradient of $F(g(x))$. I know that I should use the chain rule, but I couldn't understand chain rule in higher dimensions.

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The answer is $$(F'(g(x))\frac {\partial g} {\partial x_1},F'(g(x))\frac {\partial g} {\partial x_2},..., F'(g(x))\frac {\partial g} {\partial x_n})$$ $$=(f(g(x))\frac {\partial g} {\partial x_1},f(g(x))\frac {\partial g} {\partial x_2},..., f(g(x))\frac {\partial g} {\partial x_n})$$

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Let $h(x)=F(g(x))$. The multivariate chain rule states $dh=\frac{\partial h}{\partial dx^i}dx^i$, with implicit summation over repeated indices. By the univariate chain rule, $\frac{\partial h}{\partial x^i}=f(g(x))\frac{\partial g}{\partial x^i}$, so $dh=f(g(x))\frac{\partial g}{\partial x^i}dx^i$. In particular, $\nabla h$ is the vector whose $i$th component is the $dx^i$ coefficient, so $\nabla h=f(g)\nabla g$, in agreement with @KaviRamaMurthy's answer.