Derivative of a multivariable function evaluate the derivative of a function

Stefan Hendricks 2022-01-04 Answered
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
\(\displaystyle{g}′{\left({t}\right)}={\left({y}−{x}\right)}^{{T}}\nabla{f{{\left({x}+{t}{\left({y}−{x}\right)}\right)}}}\)

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Expert Answer

ramirezhereva
Answered 2022-01-05 Author has 1321 answers
Let \(\displaystyle{x}={\left({x}_{{1}},{x}_{{2}},\ldots,{x}_{{n}}\right)}{\quad\text{and}\quad}{y}={\left({y}_{{1}},{y}_{{2}},\ldots,{y}_{{n}}\right)}.\)
Then: \(\displaystyle{x}+{t}{\left({y}-{x}\right)}={\left[{x}_{{1}}+{t}{\left({y}_{{1}}−{x}_{{1}}\right)},\ldots,{x}_{{n}}+{t}{\left({y}_{{n}}−{x}_{{n}}\right)}\right]}.\)
So, \(\displaystyle{g{{\left({t}\right)}}}={f{{\left({x}_{{1}}+{t}{\left({y}_{{1}}−{x}_{{1}}\right)},\ldots,{x}_{{n}}+{t}{\left({y}_{{n}}−{x}_{{n}}\right)}\right)}}}.\)
Define \(\displaystyle{z}_{{i}}{\left({t}\right)}={x}_{{i}}+{t}{\left({y}_{{i}}−{x}_{{i}}\right)}.\)
So, \(\displaystyle{g{{\left({t}\right)}}}={f{{\left({z}_{{1}}{\left({t}\right)},..,{z}_{{n}}{\left({t}\right)}\right)}}}.\)
Now, \(\displaystyle{g}'{\left({t}\right)}=\sum{\frac{{\partial{f}}}{{\partial{z}_{{1}}}}}{\frac{{{\left.{d}{z}\right.}_{{1}}}}{{{\left.{d}{t}\right.}}}}\) which equals the desired product you have written.
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