Find double derivative using implicit derivation - mulitvariable calculus

Question

Short question:

$z(x,y)$ is a function that is implicitly defined by the equation

$$4x+3y+5z+4\cos(4z)+3=0$$

in the neigbourhood of the point $p=(-5\pi/16, 1/3, \pi/4)$

I am going to calculate $\partial$ in $p$.

---

$F$ differentiated with respect to $x$ is $4$ and with respect to $y$ is $3$.

Differentiated with respect to $z$ is $5-16\sin(4z)$

Then we get

$$\frac{\partial z}{\partial x}=-4/5$$

and

$$\frac{\partial z}{\partial y}=-3/5 $$

by plugging in $z=\pi/4$

But now I don't know how to calculate $\frac{\partial^2z}{\partial x \partial y}$.

Can somebody help me?

Answer 1

In computing these partials, we will assume $F_z \not = 0$. Thus, since $z = g(x,y)$, for a small neighborhood of the points $(x,y,g(x,y))$ we can choose a curve $c(t) \in \mathbb{R}^2$ such that:

\begin{align*} (x,y,g(x,y)) \in F( c(t)) \Rightarrow 0 &= \frac{d}{dt}(F \circ c)(t)\\&= \nabla F \cdot \frac{\partial}{\partial x} (x,y,g(x,y) \\&= (F_x, F_y, F_z) \cdot (1,0, g_x) \\& \Rightarrow F_x + F_z g_x =0 \\& \frac{\partial z}{\partial x} = - \frac{F_x}{F_z} \end{align*}

Here $F(x,y,z) = 4x+3y+5z+4\cos(4z)+3$. I'll leave you to the rest. Just know you should get zero.

Answer 2

For the higher derivatives you first note that $\frac{\partial z}{\partial x}(x,y) = -\frac{4}{5-16\sin(4z(x,y))}$. Now let's differentiate this expression with respect to $y$. After some work we should get that: \begin{align*} \frac{\partial^2z}{\partial x \partial y}(x,y) = -\frac{256 \frac{\partial z}{\partial y}(x,y) \cos(4z(x,y))}{(5-16\sin(4z(x,y)))^2}. \end{align*} Now remember from the first derivatives you already calculated that $\frac{\partial z}{\partial y}(x,y) = -\frac{3}{5-16\sin(4z(x,y))}$. You can now plug that into the expression for $\frac{\partial^2z}{\partial x \partial y}(x,y)$.

Hope that helped.