Why is the curve of an isothermal process above the that of a adiabatic process during compression? For an adiabatic process $pV^{\gamma}=k$ and after differentiating the equation we get that the slope of an adiabatic process is $$\frac{\mathrm dp}{\mathrm dV}=-{\gamma}\frac{\mathrm p}V$$
For an isothermal process $pV=k$ and after differentiating the equation we get that the slope of an isothermal process is $$\frac{\mathrm dp}{\mathrm dV}=-\frac pV$$
It says in my book that during compression adiabatic curve is above the isothermal curve and that during expansion the vice versa happens. My question is that since the slope of the adiabatic process is $\gamma$ times more than the slope of the isothermal process, shouldn't the curve of the former process always be above the latter.
The two cases you refer to seem to be  for the reversible adiabatic and reversible isothermal change in an ideal gas.

In an adiabatic expansion as there is no heat loss or gain to or from the surroundings ($q=0$) the energy needed for the work of expansion against the external pressure must come from the internal energy of the gas. This results in a lowering of the temperature.  

A curve of reversible adiabatic p vs. T change must therfore be below that for a reversible isothermal change. In addition as $\gamma \gt 1$, for example $5/3$ for an monoatomic gas, the adiabatic curve becomes even further below the isothermal one. The slope of the curves is not relevent in determining which is greater.