Derivative of $f(t)=\frac {1}{\rho}\log (1+\rho t)$ Could you have me to find the ferivative of $$f(t)=\frac {1}{\rho}\log (1+\rho t)$$ with repsect to $t$?
And Is it 
$$\lim_{t\to \infty} \frac {f'(t)}{t}=1?$$
Update:
Based on Hint of Surb:
$$f(t)'=\frac {1}{1+\rho t}$$
Then 
$$\lim_{t\to \infty} \frac {f'(t)}{t}=0$$
Is it correct?
 A: Hint
$$\Big(\log(u(t))\Big)'=\frac{u'(t)}{u(t)}$$
A: Notice, we have $$f(t)=\frac {1}{\rho}\log (1+\rho t)$$  Now, differentiating both the sides w.r.t. $t$ by applying chain-rule as follows
$$\frac{d}{dt}(f(t))=\frac{d}{dt}\left(\frac {1}{\rho}\log (1+\rho t)\right)$$  $$f'(t)=\frac{1}{\rho}\frac {1}{(1+\rho t)}\frac{d}{dt}(1+\rho t)$$ $$=\frac{1}{\rho}\frac {1}{(1+\rho t)}(\rho)$$   $$\color{blue}{f'(t)=\frac {1}{1+\rho t}}$$  
A: $$f(t)=\frac {1}{\rho}\ln (1+\rho t)\Longrightarrow$$
$$\frac{d}{dt}\left(\frac {1}{\rho}\ln (1+\rho t)\right)=$$
$$\frac{\frac{d}{dt}\left(\ln (1+\rho t)\right)}{\rho}=$$
$$\frac{\frac{\frac{d}{dt} (1+\rho t)}{1+\rho t}}{\rho}=$$
$$\frac{\frac{d}{dt} (1+\rho t)}{\rho (1+\rho t)}=$$
$$\frac{\frac{d}{dt} (1)+\frac{d}{dt}(\rho t))}{\rho (1+\rho t)}=$$
$$\frac{0+\rho\frac{d}{dt}(t))}{\rho (1+\rho t)}=$$
$$\frac{0+\rho\cdot 1)}{\rho (1+\rho t)}=$$
$$\frac{\rho}{\rho (1+\rho t)}=$$
$$\frac{1}{1+\rho t}$$
So:
$$\frac{d}{dt}\left(\frac {1}{\rho}\ln (1+\rho t)\right)=\frac{1}{1+\rho t}$$

$$\lim_{t \to \infty} \left(\frac{f'(t)}{t}\right)=\lim_{t \to \infty} \left(\frac{\frac{1}{1+\rho t}}{t}\right)=\lim_{t \to \infty} \left(\frac{1}{t+\rho t^2}\right)=0$$
