Partial derivatives, don't know how to solve it

Question

ok so i got this problem :

$2x \dfrac{du}{dx}+ \dfrac1{ln (x)} \dfrac{du}{dt}$, where $u=e^{\dfrac{x}{t^2}}$
I'm having trouble with understanding the first part ->$2x \dfrac{du }{dx}$
$2x \dfrac{du}{dx} = 2x \dfrac{de^{\dfrac{x}{t^2}}}{dx} = 2x *e^\dfrac{x}{t^2}*\dfrac1{t^2}$
i cheked the answer and it seems to be correct, but the question im having is why
2x is not derived ?

Answer 1

You have the expression $$2x\frac{\partial u}{\partial x}+\frac{1}{\ln x}\frac{\partial u}{\partial t}.$$ We know that $u(t,x)=\exp\left(\frac{x}{t^2}\right)$. Let's first compute its partial derivatives: $$\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\exp\left(\frac{x}{t^2}\right)=\frac{1}{t^2}\cdot\exp\left(\frac{x}{t^2}\right)$$ $$\frac{\partial u}{\partial t}=\frac{\partial}{\partial t}\exp\left(\frac{x}{t^2}\right)=\exp\left(\frac{x}{t^2}\right)\cdot\frac{\partial}{\partial t}\frac{x}{t^2}=-\frac{2x}{t^3}\exp\left(\frac{x}{t^2}\right)$$

In computing the $x$ derivative we treat $t$ as a constant, which is why it pops out in front of the exponential. Computing the $t$ derivative is a bit trickier -- we have to use the chain rule: the derivative of an exponential is an exponential, times the derivative of the inside. Here, of course, we treat $x$ as a constant.

Now we can go ahead and insert these expressions back into the original expression. The answer to your question as to why the $2x$ isn't differentiated is simply because the differential operator $\partial/\partial x$ isn't acting on it! The expression $2x\partial u/\partial x$ means "take the partial derivative of $u$ with respect to $x$ and then multiply it by $2x$." Same deal with the second term. The expression $(\partial u/\partial t)/\ln x$ means "take the partial derivative of $u$ with respect to $t$ and then divide it by $\ln x$."

In the end you should end up with the expression $$\left(\frac{2x}{t^2}-\frac{2x}{t^2\ln x}\right)\exp\left(\frac{x}{t^2}\right)=\left(1-\frac{1}{\ln x}\right)\frac{2x}{t^2}\exp\left(\frac{x}{t^2}\right),$$ if I haven't made any computational mistakes.