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I'm having trouble understanding the solution to this question which asked me to use the product rule. I understand the product rule portion but I don't understand why the last bit is what it is.

$$\frac{d}{dx} (xy)$$

The solution that is shown is: $$\begin{align} & = \frac{d}{dx}(x)y + x\frac{d}{dx}(y)\\ & = y + x\frac{d}{dx}\\ \end{align}$$

The main problem I am having with this is the last portion. Why does the $\frac{d}{dy}(y)$ not get reduced to $1$ and we are only left with $y+x$?

If someone could please help me understand this, that would be great.



Would my understanding be correct, if I said the $y$ is not reduced to $1$ because it is considered a function and $x$ is reduced to $1$ because it is the slope?

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It is not $\frac{d}{dy}y$ but $\frac{d}{dx}y$ as you first correctly wrote. Otherwise, yes, it would be $1$. – 1015 Feb 22 '13 at 1:44
I updated the question. I understand the product idea but I just don't understand the $y + x \frac{d}{dx}$ – Jeel Shah Feb 22 '13 at 1:49
@gekkostate: Please see my comment below. :) – Haskell Curry Feb 22 '13 at 1:57
up vote 4 down vote accepted

Things are clearer if you let $ y = f(x) $ and consider the function $ x \longmapsto x \cdot f(x) $. Then use the Product Rule: \begin{align} \frac{d}{dx} [x \cdot f(x)] &= \frac{d}{dx} [x] \cdot f(x) + x \cdot \frac{d}{dx} [f(x)] \\ &= 1 \cdot f(x) + x \cdot f'(x) \\ &= f(x) + x \cdot f'(x) \\ &= y + x \frac{dy}{dx}. \end{align}

We have $ \dfrac{d}{dx} [x] = 1 $, because we are interpreting $ x $ as a function, and as a linear function, $ x $ has slope $ 1 $ everywhere.

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Sorry but why did the $\frac {d}{dx} (x)$ become $1$ and not the $\frac{d}{dx} f(x)$? Is it because $y$ is a function and $x$ isn't? – Jeel Shah Feb 22 '13 at 1:51
The derivative of $ y = x $ is the slope of the function $ y = x $, which is equal to $ 1 $. When you write $ \dfrac{d}{dx} [x] $, you are differentiating $ y = x $ and not $ y = f(x) $. – Haskell Curry Feb 22 '13 at 1:52
therefore, when you are differentiating $\frac{d}{dx}(y)$ you are actually differentiating $f(x)$? – Jeel Shah Feb 22 '13 at 1:59
Yes, precisely! :) – Haskell Curry Feb 22 '13 at 2:03
Thank you so much! Thank you for being patient :) Excellently explained. – Jeel Shah Feb 22 '13 at 2:04

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