Take b=1 and a arbitrary. That shows any a in \(\displaystyle\mathbb{N}\) is hit by your function.

Jordan Mitchell

Answered 2022-01-08
Author has **2851** answers

asked 2022-01-05

Differentiation of multivariable function proof

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{v}{\left({x}\right)}}}^{{{u}{\left({x}\right)}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}={u}'{\left({x}\right)}{f{{\left({u}{\left({x}\right)},{x}\right)}}}-{v}'{\left({x}\right)}{f{{\left({v}{\left({x}\right)},{x}\right)}}}+{\int_{{{v}{\left({x}\right)}}}^{{{u}{\left({x}\right)}}}}{\frac{{\partial}}{{\partial{x}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}\)

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{v}{\left({x}\right)}}}^{{{u}{\left({x}\right)}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}={u}'{\left({x}\right)}{f{{\left({u}{\left({x}\right)},{x}\right)}}}-{v}'{\left({x}\right)}{f{{\left({v}{\left({x}\right)},{x}\right)}}}+{\int_{{{v}{\left({x}\right)}}}^{{{u}{\left({x}\right)}}}}{\frac{{\partial}}{{\partial{x}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}\)

asked 2022-01-05

Is it possible to graph out a multivariable function (3 variables at max) on a graphing calculator? If yes, how?

asked 2022-01-07

I have the multivariable function

\(\displaystyle{\log{{\left({y}^{{2}}+{4}{x}^{{2}}-{4}\right)}}}\)

and I have found the maximal domain to be

\(\displaystyle{x}^{{2}}+{\frac{{{y}^{{2}}}}{{{4}}}}{>}{1}\)

\(\displaystyle{\log{{\left({y}^{{2}}+{4}{x}^{{2}}-{4}\right)}}}\)

and I have found the maximal domain to be

\(\displaystyle{x}^{{2}}+{\frac{{{y}^{{2}}}}{{{4}}}}{>}{1}\)

asked 2022-01-04

Derivative of a multivariable function

evaluate the derivative of a function \(\displaystyle\mathbb{R}\rightarrow\mathbb{R}\) defined as

\(\displaystyle{g{{\left({t}\right)}}}={f{{\left({x}+{t}{\left({y}−{x}\right)}\right)}}}\)

where \(\displaystyle{f}:\mathbb{R}^{{n}}\rightarrow\mathbb{R}\) is a multivariable function and \(\displaystyle{x},{y}\in\mathbb{R}^{{n}}\). Prove that

\(\displaystyle{g}′{\left({t}\right)}={\left({y}−{x}\right)}^{{T}}\nabla{f{{\left({x}+{t}{\left({y}−{x}\right)}\right)}}}\)

evaluate the derivative of a function \(\displaystyle\mathbb{R}\rightarrow\mathbb{R}\) defined as

\(\displaystyle{g{{\left({t}\right)}}}={f{{\left({x}+{t}{\left({y}−{x}\right)}\right)}}}\)

where \(\displaystyle{f}:\mathbb{R}^{{n}}\rightarrow\mathbb{R}\) is a multivariable function and \(\displaystyle{x},{y}\in\mathbb{R}^{{n}}\). Prove that

\(\displaystyle{g}′{\left({t}\right)}={\left({y}−{x}\right)}^{{T}}\nabla{f{{\left({x}+{t}{\left({y}−{x}\right)}\right)}}}\)

asked 2022-01-04

First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its domain is going to be \(\displaystyle{R}^{{2}}\) and its image is going to be all real numbers, the graph of that function is defined as a subset of \(\displaystyle{R}^{{3}}\) in which the x and y axis are going to receive the inputs, and the output is going to be in \(\displaystyle{z}{\left({x},{y},{f{{\left({x},{y}\right)}}}\right)}\) ? Is that correct? Will its graph, in this example, be some kind of surface?

asked 2022-01-04

How do you define iterations of multivariable functions?

To be clear(example):

If \(\displaystyle{f}:\mathbb{R}^{{2}}\rightarrow\mathbb{R}\)

How do you define

\(\displaystyle{f}\circ{f},\text{or}{f}\circ\ldots\circ{f}?\)?

To be clear(example):

If \(\displaystyle{f}:\mathbb{R}^{{2}}\rightarrow\mathbb{R}\)

How do you define

\(\displaystyle{f}\circ{f},\text{or}{f}\circ\ldots\circ{f}?\)?

asked 2022-01-07

multivariable functions works when the partial derivatives are continous but if the function is just differentiable does the chain rule work ? I mean if \(\displaystyle{z}={f{{\left({x},{y}\right)}}}\) is differentiable function and \(\displaystyle{x}={g{{\left({t}\right)}}},{y}={h}{\left({t}\right)}\) are also differentiable functions can we write:

\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{\partial{f}}}{{\partial{x}}}}\times{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}+{\frac{{\partial{f}}}{{\partial{y}}}}\times{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)

\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{\partial{f}}}{{\partial{x}}}}\times{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}+{\frac{{\partial{f}}}{{\partial{y}}}}\times{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)