When differentiating $4x^2 + 3y^2 -3xy$ with respect to $x$, why does $3y^2$ 'disappear'?

Question

When differentiating $4x^2 + 3y^2 -3xy$ with respect to $x$, why does $3y^2$ 'disappear'?

Not too sure where this $3y^2$ goes, or why $3xy$ can turn into just $3y$. Would love an explanation behind this.

Answer 1

This is because the equation is with respect to x. Therefore, we consider $y$ and any expression with only $y$ to be a constant, which "disappears" when differentiated.

With a component with both $x$ and $y$ such as the $3xy$ in your question, we consider the $3y$ to be a constant, much like $7$ or $102$ would be considered a constant. This is also shown below: $$ 3xy=(3y)x$$ $$ \frac{d}{dx}(3y)x=(3y)\frac{d}{dx}(x)=3y$$

Answer 2

The partial derivative with respect to $x$ in a generic point $(x,y)$ is computed as \begin{align*} \frac{\partial f}{\partial x}(x,y) & = \lim_{h\to 0}\dfrac{f(x+h,y)-f(x,y)}{h}=\lim_{h\to 0}\dfrac{4(x+h)^2+3y^2-3(x+h)y - 4x^2-3y^2+3xy}{h}\\ & =\lim_{h\to 0}\dfrac{4x^2+8xh+4h^2+3y^2-3xy-3yh-4x^2-3y^2+3xy}{h} = \lim_{h\to 0}\dfrac{h(8x-3y+4h)}{h}\\ & = 8x-3y \end{align*}

In practice, this can be obtained just by differentiating with respect to $x$ while taking $y$ as a constant. In fact, considering that $y$ has some fixed value $y_0$ and denoting $g_{y_0}(x) = f(x,y_0)$, you can easily check by comparing definitions that $$ \frac{\partial f}{\partial x}(x, y_0) = g'_{y_0}(x). $$

Answer 3

This is the difference between a total and partial differential. For a total differential, you differentiate as follows:

$$d(4x^2 + 3y^2 - 3xy) \to 8x\,dx + 6y\,dy + 3x\,dy + 3y\,dx$$

However, on a partial differential, we are asking what happens when only __ is allowed to vary. If something doesn't vary, then its differential is zero (that's the definition of something not varying). Therefore, if $x$ is the only thing above allowed to vary, we can replace all $dy$s with 0. This becomes:

$$ 8x\,dx + 6y\cdot 0 + 3x\cdot 0 + 3y\,dx \to 8x\,dx + 3y\,dx $$