Electronic – Finding the impulse response from the integral

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.

The answer is in your question: we're speaking of Linear Time-Invariant systems. Their properties are linearity and time invariance.

Linearity

$$\text{if } x_1(t) \text{ produces } y_1(t) \text{ and } x_2(t) \text{ produces } y_2(t)\\ \textbf{then}\\ \alpha\cdot x_1(t) + \beta\cdot x_2(t) \text{ produces } \alpha\cdot y_1(t) + \beta\cdot y_2(t)\\ \forall\ \alpha ,\ \beta\in\mathbb{R}$$

Time invariance

$$\text{if } x(t) \text{ produces } y(t)\\ \textbf{then}\\ x(t + \tau)\text{ produces } y(t + \tau)\\ \forall\ \tau\in\mathbb{R}$$

So if you know the impulse response, and call it \$h(t)\rightleftharpoons H(f)\$, then it is clear that for an arbitrary input, if you can express it as a sum of delayed impulses, the output will be a sum of delayed impulses responses.