Walter Clyburn

2022-01-05

Differentiation of multivariable function proof

$\frac{d}{dx}{\int}_{v\left(x\right)}^{u\left(x\right)}f(t,x)dt={u}^{\prime}\left(x\right)f(u\left(x\right),x)-{v}^{\prime}\left(x\right)f(v\left(x\right),x)+{\int}_{v\left(x\right)}^{u\left(x\right)}\frac{\partial}{\partial x}f(t,x)dt$

Melinda McCombs

Beginner2022-01-06Added 38 answers

Start with

$I\left(x\right)={\int}_{v}^{u}f(t,x)dt=F(u,x)-F(v,x)$

Where F is the antiderivative of f

$\frac{d}{dx}I=\frac{d}{dx}F(u,x)-\frac{d}{dx}F(v,x)$

Now for a function$w(a\left(x\right),b\left(x\right))$ its derivative with respect to x can be written as

$\frac{\partial a}{\partial x}\left(x\right)\frac{\partial w}{\partial x}(a,\left(x\right),b\left(x\right))+\frac{\partial b}{\partial x}\left(x\right)\frac{\partial w}{\partial x}(a,\left(x\right),b\left(x\right))$

In terms of F we have$F(u,x)$ and $F(v,x)$ which remember u and v are functions of x. Therefore for each of them we can write this by plugging into the above formula. $w\left(x\right)=F\left(x\right),a\left(x\right)=v\left(x\right)$ or $u\left(x\right)$ and $b\left(x\right)=x$ .

because$\frac{\partial x}{\partial x}=1$ Now the above also holds for $F(u,x)$ so

$\frac{d}{dx}F(u,x)=\frac{\partial u}{\partial x}f(u,x)+\frac{\partial x}{\partial x}f(u,x)$

Therefore

${I}^{\prime}=f(v,x)+\frac{\partial v}{\partial x}f(v,x)-f(u,x)-\frac{\partial u}{\partial x}f(u,x)$

Now rearranging you can see that it starts to take on your form.

${I}^{\prime}={v}^{\prime}f(v,x)-{u}^{\prime}f(u,x)+f(v,x)-f(u,x)$

Now the last two terms can be written in terms of an integral.

$f(v,x)-f(u,x)={\int}_{u}^{v}{f}^{\prime}(t,x)dt$

Which then can all come together to give

$I}^{\prime}={}^{$

Where F is the antiderivative of f

Now for a function

In terms of F we have

because

Therefore

Now rearranging you can see that it starts to take on your form.

Now the last two terms can be written in terms of an integral.

Which then can all come together to give

0

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Find the point on $z=1-2{x}^{2}-{y}^{2}$ closest to $2x+3y+z=12$ using Lagrange multipliers.

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The significance of partial derivative notation

If some function like $f$ depends on just one variable like $x$, we denote its derivative with respect to the variable by:

$\frac{\mathrm{d}}{\mathrm{d}x}f(x)$

Now if the function happens to depend on $n$ variables we denote its derivative with respect to the $i$th variable by:

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}f({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n})$

Now my question is what is the significance of this notation? I mean what will be wrong if we show "Partial derivative" of $f$ with respect to ${x}_{i}$ like this? :

$\frac{\mathrm{d}}{\mathrm{d}{x}_{i}}f({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n})$

Does the symbol $\mathrm{\partial}$ have a significant meaning?The function $f(x,y,z)$ is a differentiable function at $(0,0,0)$ such that ${f}_{y}(0,0,0)={f}_{x}(0,0,0)=0$ and $f({t}^{2},2{t}^{2},3{t}^{2})=4{t}^{2}$ for every $t>0$. Define $u=(6/11,2/11,9/11)$, with the given about. Is it possible to calculate ${f}_{u}(1,2,3)$ or ${f}_{u}(0,0,0)$, or ${f}_{z}(0,0,0)$?

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So, Derivative of $f(df)$ if exists, will satisfy $limH\to 0\frac{||f(A+H)-f(A)-df(H)||}{||H||}=0$.if $F(x,y)$ and $y=f(x)$,

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1) $F(x,y)$ 𝑎𝑛𝑑 $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$

2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constantLet $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

$f(x,y)=\{\begin{array}{ll}({x}^{2}+{y}^{2})\mathrm{cos}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}},& \text{for}(x,y)\ne (0,0)\\ 0,& \text{for}(x,y)=(0,0)\end{array}$

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be

${f}_{x}(x,y)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$

which is definitely not $0$ as $(x,y)\to (0,0)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?