# Finding the gradient from the directional derivative

Hello I have the following problem #5

In order to solve it I know that $$L_f(x_o,y_o)(h_1,h_2) = f(x_o,y_o)$$ + Gradient$$f(x_o,y_o)(h_1,h_2)$$

My issue is that I am unable to find the gradient. I have tried to come up with a system of equations by stating that the directional derivative in the direction of $$u$$ is equal to the dot product of the gradient times the vector.

DirectDerivativeU(point) = Gradient$$(x_o, y_o)*(\frac{2^{1/2}}{2},\frac{2^{1/2}}{2})$$

2 = $$(f_x,f_y)*(\frac{2^{1/2}}{2},\frac{2^{1/2}}{2})$$

2 = $$\frac{2^{1/2}}{2}f_x + \frac{2^{1/2}}{2}f_y$$.

Unfortunately I am only able to come up with one equation to solve the system and find my gradient. How can I find another equation to solve this or can I please get a completely alternative method to solve this. Thank you

You are given the directional derivative in the exact direction you need it, that is, from the point $$(3,-1)$$ towards the point where you need to approximate $$f$$. So you don't need the gradient to find the directional derivative in the direction of $$\vec u$$, because you are given the value of that directional derivative.
• I prefer not to solve problems that might be homework problems, but I'll add that you do not need to find the gradient. You write that $L_f(x_o,y_o)(h_1,h_2)= \vec{\nabla} f(x_0,y_o)(h_1,h_2)$, and so you seem to want to find $L_f(x_o,y_o)(h_1,h_2)$. But you are given this in the question, where it is called ${\partial f\over \partial{\vec u}}(3,-1)$ and equals $2$. So you don't have to compute $\vec{\nabla} f(x_0,y_o)(h_1,h_2)$ to find it. – Steve Kass Mar 9 '20 at 1:53