Given $z+x\ln(z)+xe^{xy}-1=0$ find the directional derivative at $P=(0,1)$ in the direction of $v= \langle 4 \sqrt{3} , 3 \sqrt{3} \rangle$

Question

I am given the following exercise:

Given $$z+x\ln(z)+xe^{xy}-1=0$$ find the directional derivative at $P=(0,1)$ in the direction of $$v= \langle 4 \sqrt{3} , 3 \sqrt{3} \rangle$$

There's no solution on the textbook so I would like to check my reasoning.

Firstly I expressed the direction of the vector as being $v= \langle \frac{4}{5} , \frac{3}{5} \rangle$ so it is 1 unit long (I divided by its magnitude).

After that, I proceeded evaluating the partial derivatives (that's the tricky part). Using the implicit differentiation theorem and substituting the point $(0,1)$ (when $x=0$ and $y=1$ then $z = 1$)

\begin{align*} F(x,y,z) &= z+x\ln(z)+xe^{xy}-1\\ \\ F_x(x,y,z) &= \ln(z) + e^{xy}+xye^{xy} = 1\\ F_y(x,y,z) &= x^2 \cdot e^{xy} = 0\\ F_z(x,y,z) &= 1+\frac{x}{z} = 1\\ \\ \frac{\partial z}{\partial x} &= - \frac{F_x}{F_z} = -1\\ \frac{\partial z}{\partial y} &= - \frac{F_y}{F_z} = 0 \end{align*}

So the directional derivative is given by

$$D_{\vec{v}} = \langle -1 , 0 \rangle \left\langle \frac{4}{5} , \frac{3}{5} \right\rangle = - \frac{4}{5}$$

Is my solution correct? Thank you.

Answer 1

Yes, this looks fine!

Regarding some of the comments; the question could have been phrased more clearly.

Given $$z+x\ln(z)+xe^{xy}-1=0$$ find the directional derivative at $P=(0,1)$ in the direction of $$v= \langle 4 \sqrt{3} , 3 \sqrt{3} \rangle$$

The equation $z+x\ln(z)+xe^{xy}-1=0$ implicitly defines $z$ as a two-variable function (of $x$ and $y$), at least locally.

The question then asks for the directional derivative of this (implicitly defined) two-variable function in the given point and in the given direction. Your solution looks good.