Given equation V(T^n)=C . I am trying to find percentage increase in T when V is halved and n=0.5 So, method 1: using differential dT/T=(-1/n)*(dV/V). Putting dV/V=-50% I get dT/T=100%
method 2: V1T1^n=V2T2^n then T2=4T1 and that makes the increase as 300%.
Please tell me where I am mistaken?
Your original equation, put in a more readable form, is $$V\sqrt T = C$$ Where apparently $C$ is constant (btw, you need to mention such things instead of forcing your readers to figure it out by how you are using it in your calculations).
Applying the differential and using the product and chain rules, we have $$ \sqrt T dV + V\frac {dT}{2\sqrt T} = 0$$ Divide both sides by $V\sqrt T$, and you get $$\frac {dV}V + \frac {dT}{2T}= 0\\\frac {dT}{T} = -2\frac{dV}V$$
So what exactly does this tell you? It says that at all times, the rate of relative change in $T$ is $-2$ times the rate of relative change in $V$.
Note that relative change and absolute change behave differently. If $V$ goes from $V_1$ to $V_2$, while $T$ goes from $T_1$ to $T_2$, then $$\int_{T_1}^{T_2} \frac {dT}{T} = -2\int_{V_1}^{V_2} \frac {dV}{V}\\\log T_2 - \log T_1 = -2(\log T_2 - \log T_1)\\\log\left(\frac{T_2}{T_1}\right) = -2\log\left(\frac{V_2}{V_1}\right) = \log\left(\frac{V_1^2}{V_2^2}\right)\\\frac{T_2}{T_1} = \frac{V_1^2}{V_2^2}$$ Which is the long and circuitous route to finding the last relation, which follows easily from the original equation.
But you see that when you interpret what it is telling you correctly, it agrees with your second calculation: If $V_2 = \frac 12 V_1$, then $\frac{V_1^2}{V_2^2} = 4$ so $T_2 = 4T_1$.
Your problem is that rates of change have to be integrated to give you the total change.