Construct an example of a function with period 1 which is not a trig function and is differentiable at every point.

function with period 1 which is not a trig function

Any sufficiently reasonable periodic function is the sum of its Fourier series, making it a sort of trigonometric function.

If it only needs to be once differentiable, then one can use $f(x) = x^2(1-x)^2$ for $0\le x\le 1$ and then extend by periodicity to numbers not between $0$ and $1$. In other words, first let $$ g(x) = \begin{cases} x^2(1-x)^2 & \text{if }0\le x\le 1, \\[6pt] 0 & \text{otherwise}, \end{cases} $$ and then let $$ f(x) = \sum_{n=-\infty}^\infty g(x-n). $$ For each value of $x$, only one term of the sum is not $0$, so convergence is no problem.

To make it more-than-once differentiable, how about this: $$ \sum_{n=-\infty}^\infty \exp(-(x-n)^2). $$ Then work on proving the series converges and the sum is differentiable.

Based on the fractional part of a real number, define the following odd function: $f:\mathbb{R}\to\mathbb{R}$ with period $1$:

$$f(x)=\begin{cases} 1-\left(4\{x\}-1\right)^2,& 0\le\{x\}<0.5 \\ -1+\left(4\{x\}-3\right)^2,& 0.5\le\{x\}<1 \\ \end{cases}$$

Here is a plot of the function:

Now if $\{x\}<0.5$ then $\{-x\}=1-x<0.5$ and $f(x)=1-\left(4\{x\}-1\right)^2$, while $f(-x)=-1+\left(4\{-x\}-3\right)^2=-1+\left(4(1-\{x\})-3\right)^2=-1+\left(1-4\{x\}\right)^2=-f(x)$. So by symmetry, $f(-x)=f(x)$ for $\{x\}>0.5$ too. For ${x}=0.5$ have $f(x)=-1+\left(4\times0.5-3\right)^2=0$.

Also $$\lim_{\{x\}\to0^+}{f(x)}=0$$ and $$\lim_{\{x\}\to1^-}{f(x)}=0$$ So $f$ is an odd continuous function which has period $1$.
Since $x\mapsto 1-(4x-1)^2$ is differentiable on $(0,\frac{1}{2})$ and $x\mapsto -1+(4x-3)^2$ is differentiable on $(\frac{1}{2},1)$ and it is fairly straightforward to show that the one-sided derivatives agree when $\{x\}=\frac{1}{2}$ and when $x$ is in the nbhd of an integer, so $f$ is differentiable on $(-\infty,\infty)$.