## Calculus (3rd Edition)

$$u_{xx}= -\frac{1}{2} t^{-3/2}+\frac{1}{4}x^2t^{-5/2}e^{-x^2/4t}.$$
Since $u(x,t)=t^{-1/2}e^{-x^2/4t}$, by using the chain rule, we get $$u_x=t^{-1/2}e^{-x^2/4t}(-2x/4t)=-\frac{1}{2}xt^{-3/2}e^{-x^2/4t}.$$ Deriving again, we get: $$u_{xx}=-\frac{1}{2} t^{-3/2}e^{-x^2/4t}-\frac{1}{2}xt^{-3/2}e^{-x^2/4t}(-2x/4t)\\=-\frac{1}{2} t^{-3/2}+\frac{1}{4}x^2t^{-5/2}e^{-x^2/4t}.$$