What rule is this?

Question

Given u = (2x^2 + y)(x+2y^2),

du = the answer provided in the image.

My specific question is, why do we keep adding dx and dy after differentiating? For example, d(x+ 2y^2) gives (dx + 4y dy). Shouldn't it be 1 + 4y?

Why do we add dx and dy?

Answer 1

Because that's what the derivative is.

In your first calculus class you probably weren't taught to work properly with differentials, but you were taught a procedure that is basically the same as what you're doing here: "implicit differentiation".

Recall that a typical implicit differentiation problem might be

Exercise: Suppose that $x^2 + y^2 = 1$. Find the derivative of $y$ with respect to $x$.

Solution: Differentiating both sides gives $2x + 2y y' = 0$ which we solve to obtain $y' = -\frac{x}{y}$.

_{Note: The solution above is only meant on the domain where $y \neq 0$. One still needs to deal with the $y=0$ case, but that's beyond the scope of this post.}

Another relevant feature from introductory calculus that is relevant for working with differentials is the chain rule:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x}$$

For example, you would compute

$$ \frac{\mathrm{d}}{\mathrm{d}x} (2x + 1)^2 = 2 (2x+1) \cdot \frac{\mathrm{d}(2x+1)}{\mathrm{d}x}$$

The chain rule still applies to variables:

$$ \frac{\mathrm{d}y^2}{\mathrm{d}x}= 2 y \cdot \frac{\mathrm{d}y}{\mathrm{d}x}$$

Finally, the point here is that you're not computing the derivative with respect to $x$ — you're simply taking the differential itself. E.g. you're not saying $\frac{\mathrm{d}(x^2)}{\mathrm{d}x} = 2x$, you're saying $\mathrm{d}(x^2) = 2x \mathrm{d}x$

Going back to the same implicit differentiation problem above, if you were to just take the differential of both sides, it would be

$$ 2 x \mathrm{dx} + 2 y \mathrm{d}y = 0 $$

However, this equation forces $\mathrm{d}y$ to be a multiple of $\mathrm{d}x$ (except when $y=0$), so the following ratio of differentials also holds, and gives the ordinary derivatives of the sort you learned in introductory calculus:

$$ \frac{2 x \mathrm{d}x + 2 y \mathrm{d}y}{\mathrm{d}x} = 0 $$ $$ 2 x \frac{\mathrm{d}x}{\mathrm{d}x} + 2 y \frac{\mathrm{d}y}{\mathrm{d}x} = 0 $$ $$ 2 x + 2 y y' = 0 $$

(where at the end I've used $y'$ as shorthand for $\frac{\mathrm{d}y}{\mathrm{d}x}$)

Answer 2

Since y is a function of the variable x: $$\frac {d(x+ 2y^2(x))}{dx}=1+4yy' $$ Mulitply by $dx$ $$d(x+ 2y^2)=(1+4yy')dx $$ $$d(x+ 2y^2)=1dx+4yy'dx $$ $$d(x+ 2y^2)=1dx+4y\frac {dy}{dx}dx $$ Simplify last term $$d(x+ 2y^2)=1dx+4ydy $$ Or you can use this : $$df=\frac {\partial f}{\partial x}dx+\frac {\partial f}{\partial y}dy$$

Answer 3

$$u = (2x^2 + y)(x+2y^2)$$

It helps to consider $x$ and $y$ to be coordinates in $\mathbb R^2$, parameterized by a third variable, say $t$, That is think of $u$, $x$, and $y$ all as functions of $t$. Then

\begin{align} \frac{du}{dt} &= \dfrac{d}{dt}(2x^2 + y) \cdot(x+2y^2) +(2x^2 + y) \cdot \dfrac{d}{dt}(x+2y^2) \\ &= (4x\dfrac{dx}{dt} + \dfrac{dy}{dt}) \cdot(x+2y^2) +(2x^2 + y) \cdot (\dfrac{dx}{dt}+4y\dfrac{dy}{dt}) \end{align}

$\text{$``$Multiplying through$"$}$ by $dt$, you get $$du = (4x \ dx + dy) \cdot(x+2y^2) +(2x^2 + y) \cdot (dx+4y \ dy)$$

Yes, this is just a heuristic, not a proof. But it also works.

Answer 4

$d(\text{expression})=\sum_{arguments} \frac{\partial}{\partial \text{ argument}}\text{expression}\cdot d(\text{argument})$. Arguments can be any "letters" of your expression. For example if it is $4y^{2}$ then differential is obtained by taking partial derivative with respect to $y$ which is $8y$ and muplitpying it by $dy$.