Understanding gradient differentiation Questions

Question

Can anyone help me understand what they are trying to ask in these questions?

 A) Let h(x, y) = af(x, y) +bg(x, y) for real numbers a and b.
 Find hx and hy in terms of fx, fy, gx and gy. Hence or otherwise prove that
 ∇h = a∇f + b∇g.

 B) Let h(x, y) = f(x, y)g(x, y). Find hx and hy. Hence or otherwise prove
 that ∇h = f∇g + g∇f.

 c) Let h(x, y) = f(x, y)/g(x, y). Find hx and hy. Hence or otherwise prove
 that 

$\ ∇h = \frac {(g∇f − f∇g)}{g^2}$ for g(x, y) doesn't equal 0.

Answer 1

Solve for your gradient for each respective function (g(x,y) f(x,y) ) by doing this the proof part should develop itself.

Answer 2

First, remember what the gradient is: If $f$ is some function (in two variables), then

$$\nabla f(x,y) = (f_x(x,y),f_y(x,y))$$

so that the gradient is actually a vector consisting of the partial derivatives of $f$.

In e.g. part (A), they ask you do find the partial derivatives $h_x$ and $h_y$ of $h$ in terms of the partial derivatives of $f$ and $g$. Since $h(x,y) = af(x,y) + bg(x,y)$, we can first calculate the partial derivative of $h$ with respect to $x$. It becomes: $$h_x(x,y) = af_x(x,y) + bg_x(x,y).$$ This is the first entry in the gradient $\nabla h(x,y)$, and the second entry is similarly $$h_y(x,y) = af_y(x,y) + bg_y(x,y).$$ You needed to prove that $\nabla h = a\nabla f + b\nabla g$, but we've figured out that $$\nabla h(x,y) = (af_x(x,y)+bg_x(x,y),af_y(x,y)+bg_y(x,y)),$$ which I can rewrite as $$a(f_x(x,y),f_y(x,y))+b(g_x(x,y),g_y(x,y)),$$ and this last expression is exactly $a\nabla f(x,y) + b\nabla g(x,y)$. This proves part (A). I hope this makes sense to you.