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?

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
$$\frac{d}{dt}(f(t))=\frac{d}{dt}\left(\frac {1}{\rho}\log (1+\rho t)\right)$$  $$f'(t)=\frac{1}{\rho}\frac {1}{(1+\rho t)}\frac{d}{dt}(1+\rho t)$$ $$=\frac{1}{\rho}\frac {1}{(1+\rho t)}(\rho)$$   $$\color{blue}{f'(t)=\frac {1}{1+\rho t}}$$  
