# Question on "false" application of chain rule

Let's consider $$K:\mathbb{R}^n\times [0,1]\to\mathbb{R}$$ and $$f:\mathbb{R}^n\to\mathbb{R}^n$$ differentiable with $$K(u,t):=\langle f(a+t(u-a)),(u-a)\rangle=\sum\limits_{i=1}^nf_i(a+t(u-a))(u_i-a_i)$$, where $$a\in\mathbb{R}^n$$ .

Now let's take the first partial derivative (with respect to $$u_1$$) and apply the chain and product rule of differentiation:

\begin{align*} &D_1K(u,t)=D_1\left(\sum\limits_{i=1}^nf_i(a+t(u-a))(u_i-a_i)\right)\\ &=\sum\limits_{i=1}^nD_1f_i(a+t(u-a))~\underset{=?}{\underbrace{D_1(a+t(u-a))}}~(u_i-a_i)+f_1(a+t(u-a)).\end{align*}

As $$(a+t(u-a))$$ attains the form of $$\begin{pmatrix}a_1+t(u_1-a_1)\\\vdots\\a_n+t(u_n-a_n) \end{pmatrix}$$,

I thought that $$D_1(a+t(u-a))=D_1\begin{pmatrix}a_1+t(u_1-a_1)\\a_2+t(u_2-a_2)\\ \vdots\\a_n+t(u_n-a_n) \end{pmatrix}=\begin{pmatrix}t\\0\\\vdots\\0 \end{pmatrix}$$ but this doesn't match the dimension of the image of $$K$$. Where is my mistake?

Your application of the chain rule is not entirely correct. We have: $$D_1\left(f_i(a+t(u-a))\right)=\sum_jD_jf_i(a+t(u-a))D_1(a_j+t(u_j-a_j))=tD_1f_i(a+t(u-a)).$$ So, this then leads to $$D_1K(u,t)=\sum_iD_1\left(f_i(a+t(u-a))\right)(u_i-a_i)+f_1(a+t(u-a))=t\langle D_1f(a+t(u-a)),(u-a)\rangle+f_1(a+t(u-a)).$$