What are the partial derivatives of $f(2x+3y)$? Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function and $g: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be defined as
$$g(x,y) = f(2x+3y)$$
What is the partial derivative $\frac{\partial g}{\partial x}$ and what is the partial derivative $\frac{\partial g}{\partial y}$?
My try
I think it's
$$\frac{\partial g}{\partial x} = f'(2x+3y) \cdot 2$$
and
$$\frac{\partial g}{\partial y} = f'(2x+3y) \cdot 3$$
is the answer to my question. But I'm not sure about that.
 A: Here's the way I prefer to think about these things these days. The exterior derivative is
$$ \mathrm{d} g(x,y) = 2 f'(2x + 3y) \mathrm{d}x + 3 f'(2x + 3y) \mathrm{d}y $$
When you ask for the derivative with respect to $x$, that's not really what you're asking. What you're really asking is to hold $y$ constant: i.e. to set $\mathrm{d}y = 0$.

If $\mathrm{d}y = 0$, then $\mathrm{d} g(x,y) = 2 f'(2x + 3y) \mathrm{d} x$

Of course, we can now take the ratio with $\mathrm{d}x$ to get $2 f'(2x+ 3y)$ to obtain the derivative with respect to $x$ when $y$ is held constant. And this value is indeed what people mean when they write $\partial g(x,y) / \partial x$ in a context where it's implicitly understood that it means the derivative with  $y$ held constant.
A: Your answer is correct.
Let me respond to the question you posed in the comments. In order to understand the difference between $\frac{\partial g}{\partial x}$ and $\frac{dg}{dx}$, you must have an intuition. Here's the way most people think about partial and total derivatives. When taking a partial derivative, one variable is held fixed while the other changes. When taking a total derivative, changes in one variable to affect the other. 
Now, you may be wondering - what is the total derivative of a function with more than one variable? Let me work in $\mathbf{R}^n$, with local coordinates $(x_1,\cdots,x_n)$. If we can write $\omega=f_1\mathrm{d}x^1+\cdots+f_n\mathrm{d}x^n=f_I\mathrm{d}x^I$, then the derivative is:
$$\mathrm{d}{\omega} = \sum_{i=1}^n \frac{\partial f_I}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d} x^I.$$
Here, $\wedge$ is the wedge product. This is called the exterior derivative, which you will learn about in more detail later on in your education.
A: question just ask what are the partial derivatives of f(2x+3y).
So the basic is take the function in two modes.
First take the function and differentiate it in the means of 'x'. It will be your first partial derivative. It is represented as {f'(2x+3y)x}.
Secondly take the same function again and differentiate it in the means of 'y'. It will be your second partial derivative. It is represented as {f'(2x+3y)y}.
That are two parts you want to get.
let "g" be f(2x+3y)
f'(2x+3y)x=d[f(2x+3y)]/dx
f'(2x+3y)y=d[f(2x+3y)]/dy
A: The answers here are very weird. I hope this question can have a proper answer. Let $t=2x+3y$. Then the rule for differentiating is $g= f \circ t$ w/rt $z$ = $x$ or $y$ is
$$\frac{\partial g}{\partial z} = \frac{d f}{d t} \cdot \frac{\partial t}{\partial z}$$
Therefore, you are correct because you applied the rule correctly. Of course your question was 7 years ago, and you're 8,000 now.
But still the answers thus far have been very weird.
One thing that might be incorrect is the notation $f'$...or even if correct, I think it might be a little confusing or something. I mean, technically $f=f(t)$ is univariate, but $f=f(2x+3y)$ does have 2 letters there (besides $f$). I think the notation $f'$ could be at least kinda ok if you've described $f$ as $f=f(t)$ by making the substitution I did $t=2x+3y$ here.
