$f(0)=0$ ; $f(x):=\int_0^x \cos \frac 1t \cos \frac 3t \cos \frac 5t \cos \frac 7t\,dt$, $\forall x \ne 0$. Is $f$ differentiable at $0$? Let $f:\mathbb R \to \mathbb R$ be defined by $f(0)=0$ and
$$
f(x):=\int_0^x \cos \dfrac 1t \cos \dfrac 3t \cos \dfrac 5t \cos \dfrac 7t \,dt ,\quad\forall x \ne 0.
$$
Then is $f$ differentiable at $0$?
 A: First write the integrand as
$$\begin{aligned}
\cos\frac{1}{t}&\cos\frac{3}{t}\cos\frac{5}{t}\cos\frac{7}{t}\\
&=\frac{1}{8}\Bigl(1+\cos\frac{2}{t}+\cos\frac{4}{t}+\cos\frac{6}{t}+\cos\frac{8}{t}+\cos\frac{10}{t}+\cos\frac{14}{t}+\cos\frac{16}{t}\Bigr).
\end{aligned}
$$
The, let $f_k(x)=\int_0^x\cos\frac{k}{t}\,dt$ and $f_k(0)=0$. We rewrite the difference quotient for $f_k$ (for positive $x$, say, since it is even) using a substitution and an integration by parts:
$$
\begin{aligned}
\frac{f_k(x)-f_k(0)}{x}&=\frac{1}{x}\int_0^x\cos\frac{k}{t}\,dt=\bigl[u=k/t\bigr]=\frac{k}{x}\int_{k/x}^{+\infty}\frac{1}{u^2}\cos u\,du\\
&=\frac{k}{x}\Bigl[\frac{1}{u^2}\sin u\Bigr]_{k/x}^{+\infty}+\frac{2k}{x}\int_{k/x}^{+\infty}\frac{1}{u^3}\sin u\,du
\end{aligned}
$$
Taking the absolute value and estimating (triangle inequality and $|\sin u|\leq 1$), we get
$$
\Bigl|\frac{f_k(x)-f_k(0)}{x}\Bigr|\leq\frac{x}{k}+\frac{2k}{x}\int_{k/x}^{+\infty}\frac{1}{u^3}\,du=\frac{2x}{k}
$$
We find that $f_k$ is differentiable at $0$ with $f_k'(0)=0$. Since your function $f$, by the first identity, can be written as
$$
f(x)=\frac{x}{8}+\frac{1}{8}\bigl(f_2(x)+f_4(x)+f_6(x)+f_8(x)+f_{10}(x)+f_{14}(x)+f_{16}(x)\bigr)
$$
we find that your function is differentiable at $0$, and

$$f'(0)=\frac{1}{8}.$$

A: Hint:
Use
$$\begin{align}\lim_{x \to 0}\frac{1}{x}\int_0^x\cos \frac{1}{t} \,dt &= \lim_{x \to 0}\left[\frac{1}{x}\int_0^x\frac{d}{dt}\left(t^2 \sin \frac{1}t\right) \, dt- \frac{1}{x}\int_0^x2t \sin \frac{1}{t} \, dt\right] \\ &= \lim_{x \to 0}\left[x \sin \frac{1}{x}- \frac{1}{x}\int_0^x2t \sin \frac{1}{t} \, dt\right] \\ &= 0\end{align} $$
