Editing Chain rule (section)

===Case of scalar-valued functions with multiple inputs===

Let <math>f : \reals^k \to \reals</math>, and <math>g_i : \mathbb{R} \to \mathbb{R}</math> for each <math>i = 1, 2, \dots, k.</math>
To write the chain rule for the composition of functions 
<math display="block">x \mapsto f(g_1(x), \dots , g_k(x)),</math> 
one needs the [[partial derivative]]s of {{mvar|f}} with respect to its {{mvar|k}} arguments. The usual notations for partial derivatives involve names for the arguments of the function. As these arguments are not named in the above formula, it is simpler and clearer to use [[Notation for differentiation#D-notation|''D''-Notation]], and to denote by 
<math display="block">D_i f</math> 
the partial derivative of {{mvar|f}} with respect to its {{mvar|i}}th argument, and by 
<math display="block">D_i f(z)</math>
the value of this derivative at {{mvar|z}}.

With this notation, the chain rule is
<math display="block">\frac{d}{dx}f(g_1(x), \dots, g_k (x))=\sum_{i=1}^k  \left(\frac{d}{dx}{g_i}(x)\right) D_i f(g_1(x), \dots, g_k (x)).</math>

====Example: arithmetic operations====
If the function {{mvar|f}} is addition, that is, if 
<math display="block">f(u,v)=u+v,</math>
then <math display="inline">D_1 f = \frac{\partial f}{\partial u} = 1</math> and <math display="inline">D_2 f = \frac{\partial f}{\partial v} = 1</math>. Thus, the chain rule gives 
<math display="block">\frac{d}{dx}(g(x)+h(x)) = \left( \frac{d}{dx}g(x) \right) D_1 f+\left( \frac{d}{dx}h(x)\right) D_2 f=\frac{d}{dx}g(x) +\frac{d}{dx}h(x).</math>

For multiplication
<math display="block">f(u,v)=uv,</math>
the partials are <math>D_1 f = v</math> and <math>D_2 f = u</math>. Thus, 
<math display="block">\frac{d}{dx}(g(x)h(x)) = h(x) \frac{d}{dx} g(x) + g(x) \frac{d}{dx} h(x).</math>

The case of exponentiation
<math display="block">f(u,v)=u^v</math>
is slightly more complicated, as 
<math display="block">D_1 f = vu^{v-1},</math>
and, as <math>u^v=e^{v\ln u},</math>
<math display="block">D_2 f = u^v\ln u.</math>
It follows that 
<math display="block">\frac{d}{dx}\left(g(x)^{h(x)}\right) = h(x)g(x)^{h(x)-1} \frac{d}{dx}g(x) + g(x)^{h(x)} \ln g(x) \,\frac{d}{dx}h(x).</math>