1

I'm struggling with basic rules for derivatives. So $\dfrac{d}{dx} 2x = 2$ Because you factor out the constant to $2\times \dfrac{d}{dx}x$ and that is $2\times1 = 2$ But $\dfrac{d}{dx}2 = 0$ Again factor out to $2\times \dfrac{d}{dx}1(*)$ and that is $2\times0 = 0$

(*) Is that the right way to think about it that there should be a $1$?

Greetings Jeff

iostream007
  • 4,529
Jeff
  • 115

2 Answers2

2

It's certainly a way to think about it. Remember a derivative essentially tells us how much a function is changing. Constants will essentially scale how fast the function changes. $2x$ grows twice as fast as $x$, same as $2x^5$ grows twice as fast as $x^5$ and yes, $2\times 1$ grows twice as fast as $1$ but remember both $2$ and $1$ as functions don't grow at all, they're constant and hence their derivative is zero.

john
  • 5,633
  • Ah yes now I understand how I have to think about constants. Thank you. – Jeff Jun 05 '13 at 15:50
  • Derivatives are a bit confusing at first and you'll start to develop a whole bunch of rules to make them easier to deal with but the only issue with learning all of these rules is not to lose sight of what these things actually mean (this applies to a lot of mathematical concepts). – john Jun 05 '13 at 15:57
0

$$\dfrac{d}{dx}2x\implies2\dfrac{d}{dx}x\implies2\times1\implies 2$$

$\dfrac {d}{dx}C=0\;\;,C \;is\;a\;constant $ and $\;\;\dfrac{d}{dx}x=1$

for example $\dfrac{d}{dx}\ln x=\dfrac1x\;\;$but $\dfrac{d}{dx} \ln C=0$

means you are differentiating a function w.r.t. a variable and that function is not of that variable then it treated as constant so its differential is $0$

iostream007
  • 4,529