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 $(-\infty,\infty)$.

Let's go for something that is piecewise continuous.

So we need $f(0)=f(1)$ and $f'(0)=f'(1)$

A linear function will not work (well, it will, but trivially) and a quadratic function will fail on the gradient constraint, so try a cubic.

Let $f(x)=ax^3+bx^2+cx+d$.

For simplicity set $d=0$.

$f(1)=f(0)=0 \Rightarrow a+b+c=0$.

$f'(x)=3ax^2+2bx+c$

$f'(0)=f'(1) \Rightarrow c=3a+2b+c \Rightarrow 3a+2b=0 \Rightarrow b=- \frac {3a}2$.

Substitute into $a+b+c=0$ to get $a - \frac {3a}2+c=0 \Rightarrow c=\frac {a}2$.

We can scale up to avoid fractions to give $f(x)=2x^3-3x^2+x$ for $0 \le x \le 1 $ and $f(x+1)=f(x)$ for all other values of $x$.

Start from the function $x^4-x^2$. It has two symmetrical minima around the $0$. $(x^4-x^2)'=4x^3-2x=0\Leftrightarrow x=0$ or $x=\pm \frac{\sqrt 2}{2}$. At this minima the derivative is zero so you can connect the peaces between the minima smoothly. But to make it with a period of $1$ you should scale it horizontally: you want the distance between the minima to be $1$. This means you scale by $\epsilon>0:$ $g(x)=(\frac{x}{\epsilon})^4-(\frac{x}{\epsilon})^2$. Now the extremums are at $g'(x)=0\Leftrightarrow$ $x=0$ or $x=\pm \frac{\epsilon}{\sqrt 2}$. For a period of one we want $\frac{\epsilon}{\sqrt 2}-(-\frac{\epsilon}{\sqrt 2})=1\Leftrightarrow \epsilon=\frac{\sqrt 2}{2}$.

Finally, the periodic function is $f(x)=g(x-n),$ for $x\in [n-1/2,n+1/2],\,n=\pm1,\pm2,\pm3,....$ where $g(x)=4x^4-2x^2$ (plug in $\epsilon=\frac{\sqrt 2}{2}$)

Try with $\exp(\frac{-1}{1/4-|x|^2})$ for $x\in (-1/2,1/2)$, and extend it periodically.

Take any function $g(x)$ that is defined and differentiable on the domain $-1 \le x \le 2$.

Calculate $g(0)$, $g(1)$, $g'(0)$ and $g'(1)$.

Let $f(x)=(ax^2+bx+c)g(x)$

Then $f'(x)=(ax^2+bx+c)g'(x)+(2ax+b)g(x)$

$f(0)=f(1) \Rightarrow cg(0)=(a+b+c)g(1) \Rightarrow ag(1)+bg(1)+c\left[g(1)-g(0) \right]=0$

$f'(0)=f'(1) \Rightarrow cg'(0)+bg(0)=(a+b+c)g'(1)+(2a+b)g(1)\Rightarrow a\left [g'(1)+2g(1)\right]+b\left[g'(1)+g(1)-g(0)\right]+c\left[g'(1)-g'(0) \right]=0$

$b$ and $c$ can be determined from any choice of "a", so there are any number of such functions.

First, take any function $g$ that is continuous on $[0,1]$ and which verifies $g(0) = g(1)$ and $\int_0^1 g(t) dt = 0$. Extend it by periodicity on $\mathbb{R}$ and let us still call $g$ this extended function.

Then the function $x \rightarrow \int_0^x g(t) dt$ is differentiable, since it is the integral of a continuous function, and the periodicity of $g$ and the fact that the integral of $g$ is $0$ on $[0,1]$ ensures that $f$ is periodic.