Partial derivative chain rule for gradient

[physics2000]

Homework Statement

compute the gradient:

[tex] ln(z / (sqrt(x^2-y^2)) [/tex]

Homework Equations

[tex]∇=(∂/(∂x)) + ... for y and z [/tex]

I just want to know how to do the first term with respect to x

The Attempt at a Solution

I am so rusty I don't know where to begin.

 

[e^(i Pi)+1=0]

[itex]ln(z) - \frac{1}{2}ln(x^2-y^2)[/itex]

That ought to make it easier. Now treat y and z as constants.

 

[cosmic dust]

And you sum.

You don’t sum... Gradient is a vector and what you have found are the linearly independent components of it.

 

[Fredrik]

jfgobin, you shouldn't give away the final answer when you're answering posts in the homework forum. Just give a hint, like e^(i Pi)+1=0 did.
Mod note: I dealt with this.
physics2000, The hint given by e^(i Pi)+1=0 simplifies the problem significantly, but you don't have to use it. It's also possible to use these three rules directly:
\begin{align}
&\log'(x)=\frac 1 x\\
&(fg)'(x)=\frac{f'g-fg'}{g^2}\\
&\frac{d}{dx}x^a=ax^{a-1}
\end{align}

 

[jfgobin]

You don’t sum... Gradient is a vector and what you have found are the linearly independent components of it.

Yup, my bad!