# stumbled on exercise for directional derivates, $\frac{x}{\lVert r\rVert^{3/2}}$ at $(1,-1,2)$ $(r=(x,y,z))$

I'm currently self learning multivariable calculus using Serge Lang Calculus of several variables book.
yesterday night,I stumbled on of the exercise on chapter 'Directional Derivative'.
The question look straightforward, I believe I may missed something very obvious.
The exercise in question as below....

In what direction are the following functions of r increasing most rapidly at the given point?
$\frac{x}{\|r\|^\frac{3}{2}}$ at $(1,-1,2)$ $(r=(x,y,z))$

the answer in the books is given as
$$\left(\frac{9}{2.6^{7/4}},\frac{3}{2.6^{7/4}},-\frac{6}{2.6^{7/4}}\right)$$ or also (3,1,-2)

i try this as $\frac{x}{({x^2+y^2+z^2})^\frac{5}{4}}$ and then use chain rule and quotient rule but came nowhere near the answer.
I been pulling my hair all night on this.

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 Did you really mean to write $(x^2+y^2+z^2)^{5/4}$? If so, that might be the source of your difficulties since this is not consistent with the exponent in your original formulation. – Will Orrick Apr 13 '12 at 3:25 you are right now looking at it, I don't how the hell can I arrive at 5/4.I think my frusfration got better of me. – kypronite Apr 13 '12 at 3:58

Since the function is differentiable, the directional derivative in the direction $h$ is given by $df(x,h) = \langle \nabla{ f(x)} , h\rangle$; from this it can be seen that the steepest ascent directions are given by $\lambda \nabla{ f(x)}$, with $\lambda >0$. In particular, you care about the direction, not it's length. Use this fact to simplify the calculations.

The function is given by $f(x) = x (x^2+y^2+z^2)^{-\frac{3}{4}}$ (note the exponent is $-\frac{3}{4}$, not $-\frac{5}{4}$). Using the chain or quotient rules gives:

$$\frac{\partial f(x)}{\partial x} = (x^2+y^2+z^2)^{-\frac{3}{4}} - \frac{3}{2} x^2 (x^2+y^2+z^2)^{-\frac{7}{4}}$$ $$\frac{\partial f(x)}{\partial y} = - \frac{3}{2} xy (x^2+y^2+z^2)^{-\frac{7}{4}}$$ $$\frac{\partial f(x)}{\partial z} = - \frac{3}{2} xz (x^2+y^2+z^2)^{-\frac{7}{4}}$$

To simplify life, multiply through by $(x^2+y^2+z^2)^{+\frac{7}{4}}$, yielding

$$(x^2+y^2+z^2)^{+\frac{7}{4}} \frac{\partial f(x)}{\partial x} = (x^2+y^2+z^2) - \frac{3}{2} x^2$$ $$(x^2+y^2+z^2)^{+\frac{7}{4}} \frac{\partial f(x)}{\partial y} = -\frac{3}{2} x y$$ $$(x^2+y^2+z^2)^{+\frac{7}{4}} \frac{\partial f(x)}{\partial z} = -\frac{3}{2} x z$$

Now substitute the values $(x,y,z) = (1,-1,2)$, and multiply through by $\frac{2}{3}$ for simplicity, and you get $(3,1,-2)$, which was one of your answers.

 on the last part where you multiply both side with $(x^2+y^2+z^2)^\frac{7}{4}$, shouldn't it be the LHS be $(x^2+y^2+z^2)^\frac{7}{4} \frac{\partial f(x)}{\partial y}$ and not $(x^2+y^2+z^2)^\frac{5}{4} \frac{\partial f(x)}{\partial y}$ – kypronite Apr 13 '12 at 4:31 thanks, I gone through your workout and the answer produced is the same.I will be pulling my hair less tonight, many thanks :D – kypronite Apr 13 '12 at 4:47 @kypronite: Indeed it should, thanks! – copper.hat Apr 13 '12 at 5:02