Stefan Hendricks

2022-01-04

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

ramirezhereva

Let $x=\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=\left({y}_{1},{y}_{2},\dots ,{y}_{n}\right).$
Then: $x+t\left(y-x\right)=\left[{x}_{1}+t\left({y}_{1}-{x}_{1}\right),\dots ,{x}_{n}+t\left({y}_{n}-{x}_{n}\right)\right].$
So, $g\left(t\right)=f\left({x}_{1}+t\left({y}_{1}-{x}_{1}\right),\dots ,{x}_{n}+t\left({y}_{n}-{x}_{n}\right)\right).$
Define ${z}_{i}\left(t\right)={x}_{i}+t\left({y}_{i}-{x}_{i}\right).$
So, $g\left(t\right)=f\left({z}_{1}\left(t\right),..,{z}_{n}\left(t\right)\right).$
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.

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