Let $A$ be a unital $C^*$-algebra.
(a) If $a,b\in A$, show that the map $f:\mathbb{C}\to A$ defined by $f(\lambda)=e^{-i\lambda b}ae^{i\lambda b}$ is differentible and that $f^\prime (0)=i(ba-ab)$
(b) Let $X$ be a closed vector subspace of $A$ which is unitarily invariant in the sense that $uXu^*\subseteq X$ for all unitaries $u$ of $A$. Show that $ba-ab\in X$ if $a\in X$ and $b\in A$.
(c) Deduce that the closed linear span $X$ of the projections in $A$ has the property that $a\in X$ and $b\in A$ implies that $ba-ab\in X$.
Hint (a):
Note you have a product rule for differentiable maps $\mathbb C\to A$, $\frac d{dt}(f(t)\cdot g(t))=\frac{df(t)}{dt}g(t)+f(t)\frac{dg(t)}{dt}$. This product rule can be derived in the same way as the product rule for maps $\mathbb R\to\mathbb R$). The first part is then reduced to finding out what $\frac{d\exp({i\,t b})}{dt}$ is.
To find that you can use the chain rule: $$\frac{d}{dt}\exp({i\,tb})=(D\exp)(ibt)\cdot (ib)$$ here $D\exp$ is the derivative of the exponential function. What is this derivative? Here you need to look back into your analysis course to remember how to derive power series in Banach-algebras inside of their radius of convergence. Doing this will give you $D\exp=\exp$. If you plug everything in you will see the result of (a).
Hint (b):
If $b\in A$ then $e^{i\lambda b}$ is a unitary for $\lambda\in\mathbb R$. It follows from $a\in X$ that $\gamma(\lambda):=e^{i\lambda b}ae^{-i\lambda b}$ lies in $X$ for all $\lambda\in\mathbb R$. This is a path that lies entirely in the closed vector subspace $X$. Then the differential of the map $\gamma:\mathbb R\to X$ is always element of $X$, but this differential evaluated at $0$ is $i(ba-ab)$.
Hint (c):
The span of projections is unitarily invariant, since $uPu^*$ is again a projection for $P$ a projection and $u$ unitary. If you verify that the closure of this is also unitarily invariant you can use (b) to get the result.