Step response to feedback loop with disturbance

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 step is a stimulus to the system. The system is defined by the pole zero diagram irrespective of the stimulus and it looks like a 2nd order response so here's the math behind the poles: -

You have poles at co-ordinates -5 on the real part of the s-plane and +/-9 on the \$j\omega\$ of the s-plane. So you can say 5 = \$\zeta\omega_n\$ and 9 = \$\omega_n\sqrt{1-\zeta^2}\$. From this you can work out what \$\omega_n\$ and \$\zeta\$ are.

Because it looks like a 2nd order system you can use this to define the shape from the above values for \$\omega_n\$ and \$\zeta\$: -