I am not quite sure about how to prove differentiability of a function. I know by definition, if a function is differentialbe at a point, then the limit
$$\lim_{x\to a} \dfrac{f(x)-f(a)}{x-a}$$ exists.
So do we prove that limit exists by proving the left side limit equals right side limit?
Is this the only way to prove differentiability?
Then for such an integral equation
$$u(x)=x\int_o^x t^2 \cos(u(t))dt , x\in [-1,1],$$
we can first use chain rule to prove that $u'(x)=\int_o^x t^2 \cos(u(t))dt + x \cdot x^2\cos(u(x)) $ so that $u'(x)$ exists.
Then to prove that $u'$ is differentiable, can we again use the chain rule to prove that $u''(x)=x^2\cos(u(x))+3x^2\cos(u(x))-x^3\sin(u(x))\cdot u'(x) $ exists because $u'(x)$ has been proven to exist?
The last problem is how to prove that $u(x)$ is a contraction mapping? I currently always fail to get a parameter $c\in [0,1)$ When I calculate $|u(x)-u(y)|\le c|x-y|$. What I get is $\frac{4}{3}$ or $1$.
The definition is technically the "only way" to prove differntiability, but there are some results (that follows from this definition) that may be useful for you.
For instance
A product of differntiable functions is differentiable.
All polynomials are differentiable, this includes $f(x)=x$.
For a measurable function $f$, the function $F(x) = \int_0^x f(t) dt$ is differentiable whenever $f$ is continuous.
I'll leave the rest for you.