Gradient Vector of a composite function So I have a  function 
$$r= ( x^2 + y^2)^{1/2}$$
and I want to show that 
$$\operatorname{grad} f(r) = f'(r)(\operatorname{grad} r).$$
I don't really know where to begin do you say that $f(r) = (f \circ r)(x,y)$ and then use the definition of gradient to work it out. Please give a relatively basic answer as I'm new to multi-variable calculus, thanks. 
 A: Let's break it down:
\begin{align}
f &: \Re^+\to \Re\\f(z) &=\sqrt z\\f'(z)&=\frac{1}{2\sqrt{z}}\\
\\g&:\Re^2\to\Re\\ g(x,y)&=x^2+y^2\\\nabla g(x,y)&=\binom{2x}{2y}
\\\mbox{So,}
\\f'(g(x,y))&=f'(g(x,y))\cdot\nabla g(x,y)\\
&=\frac{1}{2\sqrt{x^2+y^2}}\cdot\binom{2x}{2y}
\end{align}
A: To compute the gradient of $f(r)$ you need to compute the partial derivatives. How to do it?
Use the chain rule: $f(r)$ stands for $f[(x^2+y^2)^{1/2}]$ since $r=(x^2+y^2)^{1/2}$, therefore
$$
\begin{align}
\frac{\partial}{\partial x} f(r) & = \frac{\partial}{\partial x} f[(x^2+y^2)^{1/2}] \\
 & = \frac{d}{dr} f (r) \cdot \frac{\partial}{\partial x} (x^2+y^2)^{1/2} \\
 & = f'(r) \cdot \frac{x}{(x^2+y^2)^{1/2}}.
\end{align}
$$
Likewise you obtain
$$\frac{\partial}{\partial y} f(r) = f'(r) \cdot \frac{y}{(x^2+y^2)^{1/2}}.$$
Notice that
$$\operatorname{grad}r = \left( \frac{x}{(x^2+y^2)^{1/2}}, \frac{y}{(x^2+y^2)^{1/2}} \right),$$
therefore
$$\operatorname{grad} f(r) = f'(r) \left( \frac{x}{(x^2+y^2)^{1/2}}, \frac{y}{(x^2+y^2)^{1/2}} \right) = f'(r) \cdot \operatorname{grad} r.$$
