# Find the directional derivative at a point and in the direction of a given vector.

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 $$u$$

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.
• 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 Aug 9, 2021 at 13:06
The dot product doesn't give you a vector $$(8/45,-1/15)$$ it gives you $$8/45-1/15=1/9$$.