Electronic – the relationship between a step input and an integrator

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.

Applying a sinusoidal input to, e.g., a stable, SISO, LTI system will produce an output with a transient component and a steady-state component. The transient does what all transients do - it decays to zero after a certain time. The steady-state component is a sine wave at the same frequency as the input sinusoid but, generally, with a different amplitude and also a phase shift relative to the input

It turns out (see numerous references) that the steady-state sinusoidal response (or 'frequency response') can be obtained from the system transfer function by the simple algebraic substitution: \$\small s\rightarrow j\omega\$. And hence the frequency domain is extremely accessible from the s-domain. The time domain, for example, is not nearly so easy to access as it requires the inverse Laplace transform.

To use your example, if a system has TF \$\small G(s)=\large \frac{1}{s}\$, the frequency domain gain and phase angle would be: \$\frac{1}{\omega}\$ and \$\small -\frac{\pi}{2}\$, respectively.

From a signal, rather than system, perspective, the unit step signal, \$\small H(t)\$, has the same LT as the system, above, viz: \$\small H(s)=\large \frac{1}{s}\$, and in the frequency domain this maps to the positive spectrum \$\small H(j\omega)=\large\frac{1}{j\omega}\$, and has amplitude \$\frac{1}{\omega}\$, and a phase angle of \$\small -\frac{\pi}{2}\$ relative to a notional sine wave, \$\small sin(\omega t)\$