# Let $g(t, \mathbf{x}) = t^nf(\mathbf{x})$. Find $\nabla g(t, \mathbf{x})$

I'm given that $$f: \mathbb{R}^d \rightarrow \mathbb{R}$$ is differentiable everywhere. I'm mainly confused by the notation here, I think. I'm not really sure how to take a gradient for a function defined from $$\mathbb{R}^{d+1} \rightarrow \mathbb{R}$$. I understand how take the gradient if $$g$$ were to equal $$g(\mathbf{x}) = g(x_1, \ldots, x_n)$$. But I don't really know how to start with this problem, because I'm not sure what the gradient should look like. Can someone give me some insight into how to take this gradient?

It's the same formula as usual, except that instead of your variables being named $$x^1, \ldots, x^{d + 1}$$, they are named $$t, x^1, \ldots, x^d$$: $$\nabla g(t, {\bf x}) = \left(\frac{\partial g}{\partial t}, \frac{\partial g}{\partial x^1}, \ldots, \frac{\partial g}{\partial x^n} \right) .$$
NB there is a general Leibniz (product) rule for gradients, though it doesn't save you much trouble in cases like this, i.e., of products of functions of different variables: $$\nabla(fg) = g \nabla f + f \nabla g$$.