In one test the question said $$f(x,t)=\int_{0}^{g(x,t)} e^{-u^2} du$$
Now how I can calculate $\partial^2f/\partial t^2$ ?
I have this idea:
$$\int_{0}^{g(x,t)} e^{-u^2} du=F(g(x,t))-F(0)$$
When $F$ is the primitive function of $\int e^{-u^2} du$ and then use the chain rule and FTC.
Is this approach common? What's your idea?
I should say that $g(x,t)=\frac{x}{\sqrt{2kt}}$ and so I had messy computations!
$$ \frac{\partial}{\partial t}f(x,t)=\frac{\partial}{\partial t}\int_{0}^{g(x,t)} e^{-u^2} du=e^{-g^2(x,t)}\frac{\partial}{\partial t}g(x,t)\ , $$ where one uses the fundamental theorem of calculus $\frac{d}{dz}\int_0^z dt\ h(t)=h(z).$
Taking a second derivative (using the product rule) $$ \frac{\partial^2}{\partial t^2}f(x,t)=\frac{\partial}{\partial t}[e^{-g^2(x,t)}] \frac{\partial}{\partial t}g(x,t)+e^{-g^2(x,t)}\frac{\partial^2}{\partial t^2}g(x,t)\ , $$ and I'm sure you can take it from here (use the chain rule for the first factor).