I was wondering if, and how exactly logarithmic differentiation may be applied to a multivariate function.

For example, I have been working with the function


Does the following process hold? Specifically, does the chain rule step with del f hold?


$$\Rightarrow \frac{1}{f(x,y)} \nabla f = \langle \frac{2y}{x^2-y^2} , \frac{-2x}{x^2-y^2} \rangle$$

Solving for $\nabla f$, does give me the correct answer. I'm wondering is my notation would be correct, and if this will ever not work.

Thanks for any help.

  • 1
    $\begingroup$ If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – user Jan 4 '18 at 20:17

You define


and thus

$$\nabla g=\left(\frac{1}{f(x,y)}f_x,\frac{1}{f(x,y)}f_y\right)=\frac{1}{f(x,y)}\left(f_x,f_y\right)=\frac{1}{f(x,y)}\nabla f$$

  • $\begingroup$ Ahhh, Okay, this seems very obvious now. Sorry if that was a dumb question. Thank you. $\endgroup$ – Kyle B Jan 3 '18 at 22:15
  • $\begingroup$ @KyleB You are welcome! Bye $\endgroup$ – user Jan 3 '18 at 22:22

You could take a simpler function and checked whether it holds. The usual differentiation rules hold true for the gradient as well: sum, difference, product and quotient. For instance, $\nabla(fg)=g\nabla f+f\nabla g$, $\nabla(\frac{f}{g})=\frac{g\nabla f-f\nabla g}{g^2}$, $\nabla(\ln f)=\frac{\nabla f}{f}$.


Yes I really believe your result as well as your notation are correct. May be you just need to put it in a more general setting. I would suggest you to proceed as follows.

Suppose that $f:\mathbb{R}^n\to\mathbb{R}$, which is differentiable and strictly positive in a certain open set $U\subset\mathbb{R}^n$. Then the function $F=\ln\circ f$ is also differntiable on $U$ with: $\nabla F(x)=\ln'(f(x))\cdot \nabla f(x)=\frac{1}{f(x)}\cdot\nabla f(x)$ (chain rule). Hence we can solve for $\nabla f(x)$, as we are in vector space $\mathbb{R}^n$ and get the answer right.

Remember the chain rule statement: If $G$ and $H$ are differentianble functions and $F=H\circ G$ is well defined, then $F$ is also differentiable with $\nabla F(x)=\nabla H(G(x))\cdot\nabla G(x)$.


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