Impulse response is not always a derivative of unit step response - it is the case in linear systems only. I believe your book is dealing with linear systems, therefore the method you suggest must work.
Assuming that your system is causal in addition to being linear, the unit step response is (probably) given as:
$$c(t)=(1-10e^{-t})u(t)$$
Differentiating the above equation leads to:
$$h(t)=(1-10e^{-t})\delta (t) + 10e^{-t}u(t)$$
Since Dirac's delta is zero for \$t \neq 0\$, the equation can be rewritten as:
$$h(t)=(1-10)\delta (t) + 10e^{-t}u(t)$$
Perform Laplace Transform and you'll get to the same result as your book.
Summary:
Dealing with systems don't ever forget to explicitly write the complete form of responses. Keep track of your \$\delta\$'s, \$u\$'s and etc. Differentiate the complete forms of functions.
What is the physical meaning of "first" and "second order"? ... How do I know if a system is first or second order?
A 1st order system has one energy storage element and requires just one initial condition to specify the unique solution to the governing differential equation. RC and RL circuits are 1st order systems since each has one energy storage element, a capacitor and inductor respectively.
A 2nd order system has two energy storage elements and requires two initial conditions to specify the unique solution. An RLC circuit is a 2nd order system since it contains a capacitor and an inductor
Where do equations (1) and (4) come from?
Consider the homogeneous case for the 1st order equation:
$$\tau \frac{dy}{dt} + y = 0$$
As is well known, the solution is of the form
$$y_c(t) = y_c(0) \cdot e^{-\frac{t}{\tau}}$$
which gives physical significance to the parameter \$\tau\$ - it is the time constant associated with the system. The larger the time constant \$\tau\$, the longer transients take to decay.
For the 2nd order system, the homogeneous equation is
$$\tau^2\frac{d^2y}{dt^2} + 2\tau \zeta \frac{dy}{dt} + y = 0$$
Assuming the solutions are of the form \$e^{st}\$, the associated characteristic equation is thus
$$\tau^2s^2 + 2\tau\zeta s + 1 = 0 $$
which has two solutions
$$s = \frac{-\zeta \pm\sqrt{\zeta^2 -1}}{\tau}$$
which gives physical meaning to the damping constant \$\zeta\$ associated with the system.
The transient solutions are, when \$\zeta > 1\$ (overdamped), of the form
$$y_c(t) = Ae^{\frac{-\zeta +\sqrt{\zeta^2 -1}}{\tau}t} + Be^{\frac{-\zeta -\sqrt{\zeta^2 -1}}{\tau}t} $$
when \$\zeta = 1\$ (critically damped), the solutions are of the form
$$y_c(t) = \left(A + Bt\right)e^{-\frac{\zeta}{\tau}t} $$
and when \$\zeta < 1\$ (underdamped), the solutions are of the form
$$y_c(t) = e^{-\frac{\zeta}{\tau}t}\left(A\cos \left(t\sqrt{1 - \zeta^2}\right) + B\sin \left(t\sqrt{1 - \zeta^2}\right) \right)$$
When given a first order system, why is sometimes equation (2) given,
and sometimes equation (3) as the transfer function for this system?
Different disciplines have different conventions and standard forms. Equation (2) looks to me like control theory standard while equation (3) looks like signal processing standard.
Standard forms evolve to fit the needs of a discipline. Further, if a particularly influential person or group develops and uses a particular convention, that convention often becomes the standard. It might be educational to peruse older textbooks and journals to get a sense of how notation and standard forms evolve.
Best Answer
You are probably right, but it depends on how you define the transfer function. The sine part is right, while as you can see your \$H(s)\$ is not adimensional, it's something like \$s^2\$, that is pretty strange for a transfer function[^seconds].
You are safe assuming that's a slide mistake. For the future keep in mind that checking the measurement units is always a great idea.
[^seconds]: the s is for seconds, not for the s variable.