first and second order partial derivative easy questions

It has been a long time since I've had to do derivatives so I'm rusty.

show that $\displaystyle \frac {\delta u}{\delta t} = k \frac {\delta ^2 u}{\delta x^2}$ where 'k' is constant.

$\displaystyle u=u(x,t)= exp(-n^2kt)sin(nx)$

I get :

$\displaystyle u_t(x,t)=-n^2k(1) exp(-n^2kt)sin(nx)$

and

$\displaystyle u_x(x,t)=exp(-n^2kt) ncos(nx)$

$\displaystyle u_(xx)(x,t)=exp(-n^2kt) (-n^2sin(nx))$

so if I multiply the second order derivative (U_xx) by 'k' then this should be correct?

I just want to make sure that since I'm doing partial derivatives with respect to x for the last 2 that I can ignore the $\displaystyle exp(-n^2kt)$ because it is all a cosntant. And when I do the derivative of $\displaystyle sin(nx)$ I only need to pull the 'n' out and change it to cos(nx). Also, for the first derivative with respect to 't' I am taking the '1' exponent off of 't' and bringing it forward.

Last semester was 100% integrals and I feel like I have to relearn the first semester all over. Thanks for the help.

Also, just notices that my $\displaystyle U_xx$ doesnt work. Anyone able to tell me how to get two sub x's in a row. Figured parathese would work, guess not.

Re: first and second order partial derivative easy questions

$\displaystyle \frac{\partial ^2u(x,t)}{\partial x^2} = n^2 \left(-e^{-k n^2 t}\right) \sin (n x)$

$\displaystyle \frac{\partial u(x,t)}{\partial t} = k n^2 \left(-e^{-k n^2 t}\right) \sin (n x)$

Re: first and second order partial derivative easy questions

You use braces: U_{xx} to give $\displaystyle U_{xx}$. To get the other notation, do \frac{\partial^2U}{\partial{x}^2} for $\displaystyle \frac{\partial^2U}{\partial{x}^2}$. The braces around x in the denominator are just to separate it from the word "partial". A space works, too: \frac{\partial^2U}{\partial x^2} gives the same thing.

When you take the partial derivative with respect to x, t is held constant, so $\displaystyle exp(-n^2kt)$ is a constant and you're just differentiating the sine function. The second derivative works the same way. So you go from sine to cosine to negative sine, and two factors of n come out.

When you take the partial derivative with respect to t, x is held constant, so $\displaystyle \sin{nx}$ is a constant any you're just differentiating the exponential, and a factor of $\displaystyle -n^2k$ comes out. So you have the same thing as $\displaystyle U_{xx}$ with an additional factor of k.

When I take the derivative of $\displaystyle e^{ax}$, my thinking is that I bring down the coefficient of whatever I'm differentiating with respect to. Technically, I'm using the chain rule, so I'm multiplying by the derivative of ax.

- Hollywood