Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ \mathbf v \in \Bbb R^2$ be a vector. Let $g$ be the one-variable function you get by restricting $ f(x,y)=3y^2+2xy $ to the line spanned by $\mathbf v$ and moving with constant speed 1 in direction $\mathbf v$. Find an expression for $g'(0)$.

share|cite|improve this question
up vote 0 down vote accepted

You want to find the directional derivative at $0.$ Write $v=(a,b)$ and use the fact that $D_{u+w}(f) = D_u(f) + D_w(f).$

share|cite|improve this answer
Thanks for the response! Could you elaborate on your notation on the last part of your answer? – L1meta Feb 18 '13 at 15:43
$D_v(f)$ is the derivative of $f$ in the direction of $v.$ – cats Feb 18 '13 at 21:11

Since $$Df(x,y)=\begin{bmatrix}2y & 6y + 2x\end{bmatrix},$$ $Df(0,0)$ is just zero. We have $g(t)=f(tv),$ so by the chain rule, $$ g'(t) = Df(tv) v. $$ It should be pretty obvious what $g'(0)$ is.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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