Is impulse response always differentiation of unit step response of a system? I was trying to solve a question in which the transfer function of a system was asked, its unit step response was given as:
$$c(t) = 1-10e^{-t}$$
The method that the book followed was to first find out $C(s)$ i.e. 
$$\mathcal{L}(c(t)) = \frac{1-9s}{s(s+1)}$$
Then they find out the Laplace Transform of the input i.e. $R(s) = \frac{1}{s}$ (since the input was step input)
and finally the transfer function $G(s)$ was,
$$G(s) = \frac{C(s)}{R(s)} = \frac{1-9s}{s+1} $$ 
That was the answer.
But I tried to find out the transfer function by first calculating the impulse response ($h(t)$) of the system, which is equal to the time domain differentiation of unit step response ($u(t)$).
So, 
$$h(t) = \frac{d(1-10e^{-t})}{dt} = (\delta(t))+10e^{-t}) $$
now the transfer function will be Laplace Transform of Impulse response,
So Transfer function = $1+\frac{10}{s+1}$
I can't figure out where is the mistake, why the answers differ when we apply a different approach.
 A: In the given question, system is assumed to be causal, input is unit step ,therefore step response will be $[1-10e^{-t}]u(t)$.
Now , let us apply both the methods :
1:
 $c(t) =[ 1-10e^{-t}]u(t)$
$H(s)= \frac{C(s)}{X(s)} = s\left[\frac1s-\frac{10}{s+1}\right] = \frac{1-9s}{s+1}$
2:
$c(t) = [1-10e^{-t}]u(t)$
\begin{align*}
h(t) &= \delta[c(t)]/dt  \\
 &= \delta[u(t)]/dt -10 \delta[e^{-t}u(t)]/dt)    &\text{ UV form differentiation} \\           
           &= \delta(t) + 10e^{-t}u(t) - 10e^{-t}\delta(t) \\
           &= \delta(t)-10\delta(t)+ 10e^{-t}u(t)       &                   x(t)d(t) = x(0)d(t) \\
           &= 10e^{-t}u(t) - 9\delta(t) \\
\end{align*}
$H(s) = \frac{10}{s+1} - 9 = \frac{1-9s}{s+1}$
A: 
But I tried to find out the transfer function by first calculating the impulse response of the system, which is equal to the time domain differentiation of unit step response. 

Oh, no. Just because unit impulse function is the time differentiation of unit step function, it does not follow that impulse response is the derivative of the step response. 
Instead, the step response is the convolution of unit impulse response with the step function. See here for the precise reference. So to go in the other direction, you have to find out which function gives the convolution. This is easier to do in the domain of Laplace transforms, since convolutions become multiplications, and so to go in the other direction we just have to perform division.
A: If $Y(s)$ is the impulse response, and $C(s) = \frac{1-9s}{s(s+1)}$ was the unit step input response. Its given that $c(t) = 1 - 10e^{-t}$.
$$Y(s) = sC(s)$$
Converting to time domain from s domain,
$$y(t) = \frac{d c(t)}{dt} + c(0^-)\delta(t)$$
As $$\frac{d(f(t))}{dt} <=> sF(s) - f(0^-)$$. proof
Usually, we assume zero initial conditions, so the second term is 0. i.e. $c(t = 0) = 0$.
So performing the differentiation we get,
$$y(t) = (0 +10e^{-t}) + (1-10e^0) \delta(t) = 10e^{-t} - 9\delta(t)$$
transforming this to the s domain
$$Y(s) = G(s) = \frac{10}{s+1} - 9 = \frac{1-9s}{s+1}$$

So the idea is that when you apply the step input $u(t)$ you get the output as $c(t) = u(t)(1-10e^{-t})$. Because the output can only start 
after t = 0.
$$y(t) = \frac{d c(t)}{dt} =  u(t)\frac{d (1-10e^{-t})}{dt} + (1-10e^{-t})\frac{d u(t)}{dt}$$
$$y(t) = u(t)(0 +10e^{-t}) + (1-10e^{-t}) \delta(t) $$
$$y(t) = 10u(t)e^{-t} + \delta(t) - 10\delta(t)e^{-t}  $$
Now since the dirac delta function zeros everything out (it's an impulse centered around zero) $f(t)\delta(t) = f(0)\delta(t)$
$$y(t) = 10u(t)e^{-t} + \delta(t) - 10\delta(t) $$
$$y(t) = 10u(t)e^{-t} - 9\delta(t)$$
$$Y(s) = \frac{1-9s}{s+1}$$

The following is my rough work to solve this problem.

