Show that $\nabla [f(r)]=f'(r)\frac {\mathbf{r}}{r}$

Question

Let $\mathbf{r} = xi+yj+zk$, write $r= \|\mathbf{r}\|$ and let $f:\mathbb{R}\to\mathbb{R}$ be a function of class $C^1$

So from what I know, we can derive the function at least once and we know gradients are just the derivative of the function with respect to each variable .

Anyways $$r=\sqrt{x^2+y^2+z^2}$$ now replacing $$\nabla f\left(\sqrt{x^2+y^2+z^2}\right)$$

where do I go from here to get the proof? I feel like I'm overthinking this.

The follow up is to use the answer from the above to calculate $\nabla \left(\frac{r}{\sin r}\right)$.
I am guessing $$\nabla f(r)=\nabla f\left(\frac{r}{\sin r}\right)=f'\left(\frac{r}{\sin r}\right) \frac{\|\frac{r}{\sin r}\|}{\frac{r}{\sin r}}$$

Answer 1

From $r=\sqrt{x^2+y^2+z^2}$ it follows that $${\partial r\over\partial x}={2x\over 2\sqrt{x^2+y^2+z^2}}={x\over r}\ ,$$ a formula which is extremely handy in hundreds of situations. Now you are given a function $$g(x,y,z):=f(r),\qquad r:=\sqrt{x^2+y^2+z^2}\ .$$ Using the chain rule you get $${\partial g\over\partial x}=f'(r)\>{\partial r\over\partial x}=f'(r)\>{x\over r}\ .$$ By analogy, $$\nabla g(x,y,z)=\left({\partial g\over\partial x},{\partial g\over\partial y},{\partial g\over\partial z}\right)={f'(r)\over r}\>(x,y,z)={f'(r)\over r}\>{\bf r}\ .$$

Answer 2

Remember that $D_{\mathbf r}\|v\|=\dfrac{\langle v,\mathbf r\rangle}{\|\mathbf r\|}$. Hence $D_{\mathbf r}f(\|v\|)=f'(\|\mathbf r\|)\dfrac{\langle v,\mathbf r\rangle}{\|\mathbf r\|}$. Now plug in $e_1$ $e_2$, and $e_3$ for $v$ to achieve the desired result.