Derivative of a multivariable function evaluate the derivative of a function R→R defined as g(t)=f(x+t(y−x))...

Stefan Hendricks

Answered question

2022-01-04

Derivative of a multivariable function
evaluate the derivative of a function $\mathbb{R}\to \mathbb{R}$ defined as
$g\left(t\right)=f(x+t(y-x))$
where $f:{\mathbb{R}}^{n}\to \mathbb{R}$ is a multivariable function and $x,y\in {\mathbb{R}}^{n}$. Prove that
$g\prime \left(t\right)={(y-x)}^{T}\mathrm{\nabla}f(x+t(y-x))$

Answer & Explanation

ramirezhereva

Beginner2022-01-05Added 28 answers

Let $x=({x}_{1},{x}_{2},\dots ,{x}_{n}){\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}y=({y}_{1},{y}_{2},\dots ,{y}_{n}).$ Then: $x+t(y-x)=[{x}_{1}+t({y}_{1}-{x}_{1}),\dots ,{x}_{n}+t({y}_{n}-{x}_{n})].$ So, $g\left(t\right)=f({x}_{1}+t({y}_{1}-{x}_{1}),\dots ,{x}_{n}+t({y}_{n}-{x}_{n})).$ Define ${z}_{i}\left(t\right)={x}_{i}+t({y}_{i}-{x}_{i}).$ So, $g\left(t\right)=f({z}_{1}\left(t\right),..,{z}_{n}\left(t\right)).$ Now, $g}^{\prime}\left(t\right)=\sum \frac{\partial f}{\partial {z}_{1}}\frac{{dz}_{1}}{dt$ which equals the desired product you have written.