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

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.

• There are three variables - what does $(0,1)$ mean? – Arnaud Mortier Apr 9 '18 at 14:32
• @ArnaudMortier that got me confused too but I inserted on the equation and could find $z = 1$. Is that correct? – bru1987 Apr 9 '18 at 14:37
• It does make sense, but I wouldn't say that it is correct as in "that is definitely the one thing to do". It is ambiguous. – Arnaud Mortier Apr 9 '18 at 14:39
• Maybe it should be written as $z=f(x,y)$ – John Cataldo Apr 9 '18 at 14:40
• @StanislasHildebrandt possibly. But not obvious, because of the $\ln z$ – Arnaud Mortier Apr 9 '18 at 14:40

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.