# Derivative of a multivariable function evaluate the derivative of a function

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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ramirezhereva
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.