Given $$f(x,y,z) = 2x^3y -3y^2z$$

In which directions the directional derivative at the point $a = (1,2,-1)$ is maximal (in absolute value) what is the value of the directional derivative there?

My try:

I first calculated the partial derivatives (the gradient)

$f_x = 6yx^2$
$f_y = 2x^3 -6zy$
$f_z = -3y^2$

and so $\nabla f(x,y,z) = (6yx^2, 2x^3 -6zy, -3y^2)$

Now, I need to find a direction in which the directional derivative is maximal in abs value, so I defined:
$\vec{n} = (\cos(\alpha), \cos(\beta), \cos(\gamma))$

$\frac{\partial f}{\partial \vec{n}}(1,2,-1) = \nabla f(1,2,-1) \cdot \vec{n}$

$\nabla f(1,2,-1) = (12, 14, -12)$

Now, I get a function with 3 variables! ($\alpha, \beta, \gamma$) and I am stuck as I don't know how to look for maximal points.. :

$$g(\alpha, \beta, \gamma) = 12 \cos(\alpha) +14 \cos(\beta) - 12 \cos(\gamma)$$

I don't know how to continue from here.. I would appreciate your kind help, thanks!

  • $\begingroup$ The directional derivative at some point is always maximal in the direction of the gradient at that point . It's a rather elementary result and not that hard to prove it. $\endgroup$
    – DonAntonio
    Aug 19, 2020 at 12:07

1 Answer 1


The directional derivative in the direction of a unit vector $(u,v,w)$ is $12 u+14v-12w$. By Cauchy -Schwarz inequality this is bounded by $\sqrt {(12)^{2}+(14)^{2}+(12)^{2}}$ and this is value is attained when $(u,v,w)=\frac 1 A (12,14,-12)$ where $A=\sqrt {(12)^{2}+(14)^{2}+(12)^{2}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.