I have the function: $f(x,y) = x/(x+y)$ and I want to the find the directional derivative at the point $(1,2)$ and in the direction of the vector: $a=(4,3)$.

I started by finding the gradient of $f(x,y)$ which I found to be: $(y/(x+y)^2 , -x/(x+y)^2)$, I then found the gradient at the point $(1,2)$ by substituting in $(1,2)$ and got the gradient of $f(1,2) = (2/9,-1/9)$.

Next I found the modulus of $a$ which is $\sqrt{4^2 +3^2} =5$ and I used this modulus to create $u$ where $u = a/|a|$ so $u = (4i +3j)/5$ which is equivalent to $(4/5,3/5)$.

Then from here I ran into an issue. I then tried to do $\nabla f \cdot u$ and the answer on the answer sheet is given to be $1/9$, however when I do this multiplication I get: $((4/5 \cdot 2/9), (3/5 \cdot -1/9))$ which gives me $(8/45,-1/15)$ which is not what I want.

Could someone please show me where I went wrong or what I was supposed to do once I found $\nabla f$ and $u$

Thanks in advance


2 Answers 2


Computing $\Delta f(x,y)$ we get: $$\frac{\partial f}{\partial x}(1,2)=\frac{y}{(y+x)^2}=\frac{2}{9}$$ $$\frac{\partial f}{\partial x}(1,2)=-\frac{x}{(y+x)^2}=-\frac{1}{9}$$ Then $\Delta f\cdot u$ is: $$\mathcal{D}_uf(4,3)=\frac{4}{5}\cdot \frac{2}{9}-\frac{3}{5}\cdot \frac{1}{9}=\frac{1}{9}$$ You need to add the two values, the resultant of $\Delta f\cdot u$ is not a vector.

  • $\begingroup$ Ohhh, that was my issue, for some reason I thought I had to find the modulus of my $\nabla f \cdot u$, thanks a lot for the quick answer $\endgroup$
    – Charlie P
    Aug 9, 2021 at 13:06
  • 1
    $\begingroup$ An easy mistake to make, the dot product always results in a scalar. $\endgroup$
    – benmcgloin
    Aug 9, 2021 at 13:17

The dot product doesn't give you a vector $(8/45,-1/15)$ it gives you $8/45-1/15=1/9$.


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