Find the directional derivative of $f_{\vec u}(5,2)$ for the function $f(x,y)=xy+y^3$ with $\frac{(3i-4j)}{5}$?
I got $-\frac{17}{5}$ (as the answer) from the derived formula below.
$ \left(2\right)\left(\frac{3}{5}\right)+\left(5\right)+3\left(2\right)^2\left(-\frac{4}{5}\right)$
Incase the above is not readable here is the problem on symbolab.
It is in the form $ai+bj$ where $a=y$ and $b=x+3y^2$, as can be seen in the Symbolab, I already plugged in the $x$ and $y$ values which is how I got to my final answer of $-\frac{17}{5}$.
$a$ and $b$ were found by taking the partial derivative of $x$ and $y$ via Symbolab.
Obviously my answer was wrong, I am not sure where I went wrong.
Best Answer
Directional derivative is obtained by taking the dot product of the gradient with the unit vector in the direction of the vector.
$\hat u= \frac{3}{5}\hat i - \frac{4}{5}\hat j$ is a unit vector
$\nabla U=\frac{\partial U}{\partial x}\hat i + \frac{\partial U}{\partial y}\hat j$
$U=xy+y^3$
$\frac{\partial U}{\partial x}=y \implies \frac{\partial U}{\partial x}|_{x=5,y=2}=2$
$\frac{\partial U}{\partial y}=x+3y^2 \implies \frac{\partial U}{\partial x}|_{x=5,y=2}=5+3(4)=17$
$\nabla U |_{x=5,y=2}=2\hat i+17 \hat j$
$D_u=(2 \times\frac{3}{5})+(17 \times \frac{-4}{5})=\frac{-62}{5}$