How would I approach this?

Question

I am stuck on a homework problem that involves using/taking partial derivatives with respect to x. [This is a sample problem][1]

$$ \frac{d}{dx}e^{-\large\frac{(x-y)^2}{2z^2}}$$

I really need help approaching this problem as I dont think im doing this correctly. Here is my approach:

1. Use Quotient rule on the numerator of the e-power
2. Treat the denominator as a constant and move it out of the derivation
3. solving for the derivative of the result from the quotient rule

But I went to two tutors and they gave me 2 different answers.

1. You can treat the denominator as 0
2. You can distribute the numerator using the quotient rule and then move the denominator out.

I went with the 2nd answer

Answer 1

To compute the derivative you stated just treat the variables $y, z$ as constants and differentiate $f(x,y,z) = e^{-\frac{(x - y)^2}{2z^2}}$ with respect to $x$:

$$ \partial_xf(x,y,z) = -e^{-\frac{(x - y)^2}{2z^2}}\frac{(x - y)}{z^2} $$

Answer 2

This may be a bit confusing because the function that you want to differentiate has multiple letters that are usually variables. We are just differentiating with respect to $x$ however, so we can treat this as a single-variable function. We can let $f(x)=e^{-\frac{(x-y)^2}{2z^2}}$. Notice that this function only depends on $x$. The other letters $y,z$ are not changing. To differentiate the function, we can just treat the other letters as constants.

So, to differentiate with respect to $x$. We can treat $y$ and $z$ as constants, just like you would normally do when differentiating $5x$ for example.

Now, we know that $$ \frac{d}{dx} e^{u(x)}= e^{u(x)}\cdot \frac{d}{dx}u(x) \tag{1} \label{1} $$

Let us first look at $u(x)$ in this case. We have $u(x)=-\frac{(x-y)^2}{2z^2}$ where we treat $z,y$ as constants. Differentiating this with respect to $x$ results in: $$ u'(x)=\frac{1}{2z^2}\cdot \frac{d}{dx}-(x-y)^2 $$. We can use the chain rule to differentiate this: $$ u'(x)=-\frac{1}{2z^2}\cdot 2(x-y) =-\frac{x-y}{z^2} $$ We can plug this into \eqref{1} to get: $$ \frac{d}{dx}e^{\frac{-(x-y)^2}{2z^2}} = -e^{\frac{-(x-y)^2}{2z^2}} \cdot \frac{x-y}{z^2} $$

Usually we denote functions that have multiple variables as $f(x,y,z)$ (For example). In this case the letters $y,z$ could change. Differentiating functions like these is possible as well. You can look at https://en.wikipedia.org/wiki/Partial_derivative to find out more!