# How to find the gradient of $x+y-z-\ln z=0$?

I need to find the directional derivative of $x+y-z=\ln z$ in the point $(1,0,1)$ in the direction which is formed at the angle of $\arctan({4 \over 3})$ with the positive values of $x$ axis.

The gradient $\nabla f=\langle f_x, f_y,f_z\rangle=\langle z_x, z_y,z_z\rangle$.

I got the partial derivative for $z_x=z_y=\frac{z}{z+1}$ but I don't understand how to differentiate the given function implicitly for $z$.

Any suggestions would be highly welcome.

• You want to define $f = x + y - z - \ln z \implies \nabla f = (1, 1, -1 - 1/z)$ – Mattos May 18 '17 at 6:13
• @Mattos can you please explain how you got there? I guess my calculation is not correct – Yos May 18 '17 at 6:16
• You haven't specified a vector direction to go in. So either you only need to calculate $\nabla f$ or you haven't provided all the information in the question. – Mattos May 18 '17 at 6:18
• @Mattos I added more details. I thought we can calculate the directional derivative by $||\nabla f||\cdot \cos \theta$. But first I need $||\nabla f||$ – Yos May 18 '17 at 6:21
• @Mattos I actually think I got confused with finding the gradient in 4 dimensions, so in our case $\nabla f=\langle f_x, f_y \rangle$ right? – Yos May 18 '17 at 6:33