# 6. Differential calculus of multivariate functions

### Definition 1:

Let $$X$$ and $$Y$$ be normed spaces. A mapping $$f: E \rightarrow Y$$ of a set $$E \subset X$$ into $$Y$$ is differentiable at an interior point $$x \in E$$ if there exists a continuous ***linear transformation*** $$L(x): X \rightarrow Y$$ such that

$$
f(x+h)-f(x)=L(x) h+\alpha(x ; h)
$$

where $$\alpha(x ; h)=o(h)$$ as $$h \rightarrow 0, x+h \in E$$.

We often denote that $$L(x)$$as $$\mathrm{d} f(x), D f(x)$$, or $$f^{\prime}(x)$$.

> remark: If the condition for differentiability of the mapping $$f$$ at some point $$a$$ is written as
>
> $$f(x)-f(a)=L(x)(x-a)+\alpha(a ; x)$$where $$\alpha(a;x) = o(x − a)$$ as $$x \to a$$.

### tangent space

We remark that strictly speaking $$f^{\prime}(x) \in \mathcal{L}\left(T X\_{x} ; T Y\_{f(x)}\right)$$ here and $$h$$ is a vector of the tangent space $$T X\_{x}$$. But parallel translation of a vector to any point $$x \in X$$ is defined in a vector space, and this allows us to identify the tangent space $$T X\_{x}$$ with the vector space $$X$$ itself. 

Consequently, after choosing a basis in $$X$$, we can extend it to all the tangent spaces $$T X\_{x}$$. This means that if, for example, $$X=\mathbb{R}^{m}$$,  $$Y=\mathbb{R}^{n}$$,and the **mapping** $$f \in \mathcal{L}\left(\mathbb{R}^{m} ; \mathbb{R}^{n}\right)$$ is given by the matrix $$\left(a\_{ij}\right),$$ then at every point $$x \in \mathbb{R}^{m}$$ the **tangent mapping** $$f^{\prime}(x): T \mathbb{R}*{x}^{m} \rightarrow T \mathbb{R}*{f(x)}^{n}$$ will be given by the same matrix.

### Chain Rule

If the mapping $$f: U \rightarrow V$$ is differentiable at a point $$x \in U \subset X$$, and the mapping $$g: V \rightarrow Z$$ is differentiable at $$f(x)=y \in V \subset Y$$, then the composition $$g \circ f$$ of these mappings is differentiable at $$x$$, and

$$
(g \circ f)^{\prime}(x)=g^{\prime}(f(x)) \circ f^{\prime}(x)
$$

### Differentiation of the Inverse (#)

Let $$f: U \rightarrow Y$$ be a mapping that is continuous at $$x \in U \subset X$$ and has an inverse $$f^{-1}: V \rightarrow X$$ that is defined in a neighborhood of $$y=f(x)$$ and continuous at that point. If the mapping $$f$$ is differentiable at $$x$$ and its tangent mapping $$f^{\prime}(x) \in \mathcal{L}(X ; Y)$$ has a continuous inverse $$\left\[f^{\prime}(x)\right]^{-1} \in \mathcal{L}(Y ; X),$$ then the mapping $$f^{-1}$$ is differentiable at $$y=f(x)$$ and

$$
\left\[f^{-1}\right]^{\prime}(f(x))=\left\[f^{\prime}(x)\right]^{-1}
$$

### Partial derivative

Let $$U=U(a)$$ be a ***neighborhood*** of the point $$a \in X=X\_{1} \times \cdots \times X\_{m}$$ in the direct product of the normed spaces $$X\_{1}, \cdots, X\_{m}$$ , and let $$f: U \rightarrow Y$$ be a mapping of $$U$$ into the normed space $$V$$. In this case

$$
y=f(x)=f\left(x\_{1}, \ldots, x\_{m}\right)
$$

and hence, if we fix all the variables but $$x\_{i}$$ *in the above equation* by setting $$x\_{k}=a\_{k}$$ for $$k \in {1, \ldots, m} \backslash i,$$ we obtain a function

$$
f\left(a\_{1}, \ldots, a\_{i-1}, x\_{i}, a\_{i+1}, \ldots, a\_{m}\right)=: \varphi\_{i}\left(x\_{i}\right)
$$

defined in some neighborhood $$U\_{i}$$ *of* $$a\_{i}$$ in $$X$$.

We usually denote this partial derivative by one of the symbols

$$
\partial\_{i} f(a), \quad D\_{i} f(a), \quad \frac{\partial f}{\partial x\_{i}}(a), \quad f\_{x\_{i}}^{\prime}(a)
$$

In accordance with these definitions $$D\_{i} f(a) \in \mathcal{L}\left(X\_{i} ; Y\right) .$$ More precisely, $$D\_{i} f(a) \in \mathcal{L}\left(T X\_{i}\left(a\_{i}\right) ; T Y(f(a))\right)$$.

Denote that $$h=\left(h\_{1}, \ldots, h\_{m}\right) \in T X\_{1}\left(a\_{1}\right) \times \cdots \times T X\_{m}\left(a\_{m}\right)=T X(a)$$, we have the equation

$$
\mathrm{d} f(a) h=\partial\_{1} f(a) h\_{1}+\cdots+\partial\_{m} f(a) h\_{m}
$$

### partial derivative of higher order

Consider  $$U=U(x)$$ be a ***neighborhood*** of the point $$x=(x^1,\cdots,x^n)$$, Then if $$f$$has partial derivative with respect to $$x^i$$, then $$\frac{\partial f}{\partial x^{i}}$$becomes a new function which is defined in $$U$$, if:

$$
\frac{\partial}{\partial x^{i\_{1}}}\left(\frac{\partial}{\partial x^{i\_{2}}} \cdots\left(\frac{\partial f}{\partial x^{i\_{k}}}\right) \cdots\right)(x)
$$

exist, we call it ***higher-order partial derivative***, often denoted as:

$$
\frac{\partial^{k} f}{\partial x^{i\_{1}} \cdots \partial x^{i\_{k}}}(x)
$$

## **Taylor's theorem**

If a mapping $$f: U \rightarrow Y$$ from a neighborhood $$U=U(x)$$ of a point $$x$$ in a normed space $$X$$ into a normed space $$Y$$ has derivatives up to order $$n-1$$ inclusive in $$U$$ and has an nth order derivative $$f^{(n)}(x)$$ at the point $$x$$, then

$$
f(x+h)=f(x)+f^{\prime}(x) h+\cdots+\frac{1}{n !} f^{(n)}(x) h^{n}+R\left(x;h\right)
$$

$$R\left(x;h\right)= o(|h|^n)$$when $$h\to 0$$.

We can expand the $$f^{(n)}(x)$$, such as **Jacobian matrix**: 

$$
\mathbf {T}f(x)= \mathrm df(x)=\nabla f(x)=\left({\frac {\partial f}{\partial x\_{1}}}(x),{\frac {\partial f}{\partial x\_{2}}}(x),\cdots ,{\frac {\partial f}{\partial x\_{n}}}(x)\right)
$$

or ***Hessian matrix***:

$$
\mathbf{H} f(x)=\left(\begin{array}{cccc}
\frac{\partial^{2} f}{\partial x\_{1}^{2}} & \frac{\partial^{2} f}{\partial x\_{1} \partial x\_{2}} & \cdots & \frac{\partial^{2} f}{\partial x\_{1} \partial x\_{n}} \\&&\\
\frac{\partial^{2} f}{\partial x\_{2} \partial x\_{1}} &
\frac{\partial^{2} f}{\partial x\_{2}^{2}} & \cdots & \frac{\partial^{2} f}{\partial x\_{2} \partial x\_{n}} \\&&\\
\vdots & \vdots & \ddots & \vdots \\&&\\
\frac{\partial^{2} f}{\partial x\_{n} \partial x\_{1}} & \frac{\partial^{2} f}{\partial x\_{n} \partial x\_{2}} & \cdots & \frac{\partial^{2} f}{\partial x\_{n}^{2}}
\end{array}\right)
$$

$$f^n(x)$$means all the $$n$$-th order partial derivatives of $$f$$

$$
f^{(n)}(x)h^n=\sum\_{\alpha *{1}+\cdots+\alpha *{n}=n}{\frac {\partial ^{n}f}{\partial x*{1}^{\alpha *{1}}\cdots \partial x*{n}^{\alpha* {n}}}}(x)\cdot h\_1^{\alpha\_1}\cdots h\_n^{\alpha\_n}
$$

and

$$
R\left(x;h\right)=\frac{1}{n !} \int\_{0}^{1}(1-t)^{n} D^{n+1} f(x+t\cdot h)) \mathrm d t
$$

## Lagrange multiplier <a href="#firstheading" id="firstheading"></a>

... pass.

$$\blacksquare$$


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