A body is dropped in a well and it travels $p$ depth in $t$ time interval. The relation is $$p(t)=\frac4{4+t^2}+0.8t-1.$$ Find the velocity and acceleration.
Now if I differentiate the expression by considering "$-1$" as a constant, it vanishes and velocity, $v=\dfrac{(8t)}{(4+t^2)^2} +0.8$ Again if I simplify the 1st term with "$-1$" the velocity, $v=\dfrac{-(8t)}{(4+t^2)^2}$. The book considers the later one as answer. why?
Your velocity expression is slightly incorrect,
recall that $$\frac{d}{dx} \frac{u(x)}{v(x)}=\frac{u'(x) v(x)-v'(x) u(x)}{(v(x))^2}$$
So, evaluating $$\frac{d}{dt} \frac{4}{4+t^2}=\frac{0-8t }{(4+t^2)^2}$$ So now you can see where the '$-$' sign comes from.