The function $z(x,y)$ is defined near $(1,1,0)$ by the equation:
$3\sin(x^2yz)-3x+2y^2+e^{yz}=0$.
In the direction which the value of the directional derivative of $z$ in the point $(1,1)$ is maximal, what is the value of the derivative?
I know that according to the implicit function theorem, if $z(x,y)$ is defined near $(1,0,0)$
That means $z_x,z_y$ are both continuous which means $z(x,y)$ is differentiable.
And if $z$ is differentiable there, that means that the directional derivative will have a maximum value whenever we pick the direction of the gradient vector $\nabla z=(\frac{F_x}{F_z},\frac{F_y}{F_z})$.
$$F_x=3\cos(x^2yz)[2xyz + x^2yz_x] – 3 + e^{yz}yz_x |_{(1,1,0)}=3z_x-3+z_x=4z_x-3$$
But as you see, I'm struggling with the derivatives, and I don't know how to deal with $z_x$.
Would appreciate any feedback, thanks in advance.
Note: Final answer is $\frac{5}{4}$
Best Answer
$3\sin(x^2yz)-3x+2y^2+e^{yz}=0$
$3 \cos (x^2 yz) (x^2y\frac{\partial z}{\partial x} + 2 xyz) - 3 + y e^{yz} \frac{\partial z}{\partial x} = 0 $
At $(1, 1, 0)$,
$ 3 \frac{\partial z}{\partial x} - 3 + \frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = \frac{3}{4}$
Similarly show that $\frac{\partial z}{\partial y} = - 1 $
$\nabla z = \left(\frac{3}{4}, -1\right)$
That will give you $|\nabla z| = \frac{5}{4}$
Can you take it from here?