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Is the quotient rule applicable when dealing with differetiating functions of vectors?

Pete
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1 Answers1

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The expression to be differentiated has to make sense in the first place; e.g., it should not be a quotient of vectors. Assume this is the case. For instance, we are given a function

$$h: \ {\mathbb R}^n\ \to\ {\mathbb R}, \quad h(x)\ := {f(x)\over g(x)}$$

with real-valued $f$ and $g$, and we want to compute its gradient $\nabla h$. The $i$th component of this gradient is obtained by differentiating the quotient ${f\over g}$ with respect to the single real variable $x_i$, and for this operation the quotient rule is certainly valid:

$$\bigl(\nabla h\bigr)_i(x)={\partial \over \partial x_i}\Bigl( {f(x)\over g(x)}\Bigr)= {f_{.i}(x)g(x)-f(x)g_{.i}(x) \over g^2(x)}\qquad(1\leq i\leq n)\ .$$

In vectorial notation we therefore have

$$\nabla h(x)={g(x)\ \nabla f(x)\ -\ f(x)\ \nabla g(x) \over g^2(x)}\ ,$$

or

$$\nabla{f\over g}={g\ \nabla f\ -\ f\ \nabla g\over g^2}\ ,$$

which looks precisely like the familiar quotient rule.

Christian Blatter
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  • Are we asking that $f$ and $g$ are continuously differentiable? Where was this used when computing the gradient of $f/g$? – Danilo Gregorin Afonso Mar 21 '20 at 19:41
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    @DaniloGregorin: This is about the vector $\nabla h(p)$ at some point $p$ in the domain of $h$. Wether $h$ is differentiable at $p$ is another matter. The assumptions $f$ and $g$ differentiable at $p$, and $g(p)\ne0$ are sufficient. – Christian Blatter Mar 21 '20 at 19:48