Evaluate $\nabla (3r^2-4\sqrt{r}+\frac{6}{\sqrt[3] r})$

Question

Im trying to do this excercise from Schaum's vector analysis i cant solve it but this is my attempt

Let $r=\sqrt {x^2+ y^2+z^2}$

Sust. we have

3$\left(\sqrt{x^2+y^2+z^2}\right)^2-4\sqrt {\sqrt{x^2+y^2+z^2}}+\frac{6}{\sqrt[3]{x^2+y^2+z^2}}$

And doing the partials and some algebra I got

$6(x^2+y^2+z^2)x\hat i-2(x^2+y^2+z^2)^\frac{-3}{4}y\hat j-18(x^2+y^2+z^2)^\frac{-5}{2}z\hat k$

So that

($6r-2(r)^\frac{-3}{4}-18(r)^\frac{-5}{2}$)$(x\hat i+y\hat j+z\hat k)$

Hence

($6r-2(r)^\frac{-3}{4}-18(r)^\frac{-5}{2})~~\vec r$

But the answer is

$\left(6-2(r)^\frac{-3}{2}-2(r)^\frac{-7}{3}\right) ~~\vec r$

So I don't see where I made a mistake.

Answer 1

• Hint: using the identity (which you might be able to easily prove), $$ \nabla f(r)= \frac{\vec{r}}{r} ~ \frac{\partial} {\partial r} f(r) , $$ can you evaluate a simple derivative of your function?