数学代写|MATH4604 Calculus

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MATH4604课程简介

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Prerequisites 

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MATH4604 Calculus HELP（EXAM HELP， ONLINE TUTOR）

To prove this statement, we need to show that the directional derivative of $L\circ g$ at $u$ in the direction $v$ is equal to the composition of the directional derivative of $g$ at $u$ in the direction $v$ with $L$, for any $v\in U$.

By definition, we have:

$$D_u(L\circ g)(v) = \lim_{t\to 0} \frac{(L\circ g)(u+tv) – (L\circ g)(u)}{t}.$$

Since $L$ is a linear transformation from $X$ to $W$, we can write $(L\circ g)(u+tv) = L(g(u+tv))$, and $(L\circ g)(u) = L(g(u))$. Substituting these expressions, we obtain:

$$D_u(L\circ g)(v) = \lim_{t\to 0} \frac{L(g(u+tv)) – L(g(u))}{t}.$$

Next, we can use the definition of the directional derivative again, this time for $g$, to write:

$$D_u g(v) = \lim_{t\to 0} \frac{g(u+tv) – g(u)}{t}.$$

Note that $g(u+tv)$ is an element of $W$, so we can apply $L$ to it. Using linearity of $L$ again, we get:

$$L(g(u+tv)) – L(g(u)) = L(g(u+tv) – g(u)).$$

Substituting this expression back into the equation for $D_u(L\circ g)(v)$, we obtain:

$$D_u(L\circ g)(v) = \lim_{t\to 0} \frac{L(g(u+tv) – g(u))}{t} = L\left(\lim_{t\to 0} \frac{g(u+tv) – g(u)}{t}\right) = L(D_u g(v)).$$

Therefore, we have shown that $D_u(L\circ g)(v) = L(D_u g(v))$ for any $v\in U$, which is equivalent to $D_u(L\circ g) = {}^*L\circ(D_u g)$, where ${}^*L$ is the extension of $L$ to a map from $\mathrm{TS}$ to $\mathrm{TW}$, defined by ${}^*L(v) = L\circ v$ for any $v\in \mathrm{TS}$.

To show that $\partial_i(L\circ f) \supseteq L\circ (\partial_i f)$, we need to show that for any $v\in V$, if $v\in \mathrm{dom}(\partial_i f)$ and $D_v f_i = w\in W$, then $D_v (L\circ f)j \geq (L\circ \partial_i f){j}(v)$ for any $j\in X$.

By definition of the partial derivative, we have:

$$\partial_i f(v) = D_v f_i,$$

and

$$\partial_i(L\circ f)(v) = \lim_{t\to 0} \frac{(L\circ f)(v+it\mathbf{e}_i) – (L\circ f)(v)}{t}.$$

Since $L$ is a linear transformation from $X$ to $W$, we can write $(L\circ f)(v+it\mathbf{e}_i) = L(f(v+it\mathbf{e}_i))$, and $(L\circ f)(v) = L(f(v))$. Substituting these expressions, we obtain:

$$\partial_i(L\circ f)(v) = \lim_{t\to 0} \frac{L(f(v+it\mathbf{e}_i)) – L(f(v))}{t}.$$

By the definition of $D_v (L\circ f)_j$, we have:

$$D_v (L\circ f)j = \lim{t\to 0} \frac{(L\circ f)_j(v+it\mathbf{e}_i) – (L\circ f)_j(v)}{t}.$$

Now, we can use the chain rule for directional derivatives to write:

$$(L\circ f)_j(v+it\mathbf{e}_i) = L(f(v+it\mathbf{e}_i))j = (L\circ D{v+it\mathbf{e}_i}f)_j.$$

Using the linearity of $L$ and the definition of the partial derivative, we can further simplify:

$$(L\circ D_{v+it\mathbf{e}_i}f)j = L(D{v+it\mathbf{e}_i}f)_j = L(\partial_i f(v+it\mathbf{e}_i))_j.$$

Substituting this expression back into the equation for $D_v (L\circ f)_j$, we obtain:

$$D_v (L\circ f)j = \lim{t\to 0} \frac{L(\partial_i f(v+it\mathbf{e}_i))_j – L(\partial_i f(v))_j}{t} = (L\circ \partial_i f)_j(v).$$

Therefore, we have shown that $D_v (L\circ f)j \geq (L\circ \partial_i f){j}(v)$ for any $j\in X$, which implies that $\partial_i(L\circ f) \supseteq L\circ (\partial_i f)$ as desired.

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