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SolomonZelman
 one year ago
uncommon or complicated derivatives
SolomonZelman
 one year ago
uncommon or complicated derivatives

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8At first, the definition of the absolute value is: \(\large\color{slate}{ \displaystyle \left~B~\right =\sqrt{B~^2~} }\)  So we have a function \(\large\color{slate}{ \displaystyle y=\left~f(x)~\right }\) And we would like to differentiate it. We would use the definition. \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left~f(x)~\right = }\) \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\sqrt{~f(x)~^2}~\right) = }\) (I apply the definition of absolute value to the f(x). ) We know that the derivative of a square root of x is \(\large\color{slate}{ \displaystyle \frac{1}{2\sqrt{x}} }\) This way, we get: \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\sqrt{~f(x)~^2}~\right) =\frac{1}{2\sqrt{ f(x)^2~}} }\) and we need the chain rule for the part inside the square root. \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\sqrt{~f(x)~^2}~\right) =\frac{1}{2\sqrt{ f(x)^2~}} \times ~\left[\frac{d}{dx}\left(~f(x)^2~\right) ~\right] }\) and now, we need another chain rule (for cases when f(x) is not simply an x) \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\sqrt{~f(x)~^2}~\right) =\frac{1}{2\sqrt{ f(x)^2~}} \times ~\left[\frac{d}{dx}\left(~f(x)^2~\right) ~\right] \times f'(x) }\) Now, we can simplify this whole "mess" First simplification is that we will finish differentiating the \(f(x)^2\) part. \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\sqrt{~f(x)~^2}~\right) =\frac{1}{2\sqrt{ f(x)^2~}} \times ~\left[2\times f(x) ~\right] \times f'(x) }\) now, we convert the bottom, the \(2\sqrt{f(x)^2~}\) using our definition and our definition was: \(\large\color{slate}{ \displaystyle \left~B~\right =\sqrt{B~^2~} }\) \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\leftf(x) \right~\right) =\frac{1}{2\leftf(x) \right } \times ~\left[2\times f(x) ~\right] \times f'(x) }\) now algebra: \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\leftf(x) \right~~\right) =\frac{2f(x) }{2\leftf(x) \right } \times f'(x) }\) \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~~\leftf(x) \right~\right) =\frac{f(x) }{\leftf(x) \right } \times f'(x) }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8So, you can take home the following formula. \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left~f(x)~\right =\frac{f(x) }{\leftf(x) \right } \times f'(x) }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8I will be adding more stuff later on.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You should have wrote:\[ \frac{d}{dx} \bigg( f(x)^2 \bigg) = \frac{d}{d(f(x))} \bigg( f(x)^2 \bigg) f'(x) \]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8where is the square root ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8d(f(x)).... I thought it would be more understandable to read d/dx ( ...) to be like "oh I am differentiating with respect to x"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \begin{split} \frac{d}{dx}\left(\begin{cases} x &x>0\\ x &x<0 \end{cases} \right) &= \begin{cases} 1 &x>0\\ 1&x<1 \end{cases} \\ &=\begin{cases} \text{sgn}(x) &x \neq 0\\ \text{undef} \end{cases} \\ &=\frac{x}{x} &= \frac{x}{x} \end{split} \]Once you have this, chain rule can be used for \(f(x)\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The problem is that \[ \frac{d}{d(f(x))} f(x)^2 = 2f(x) \]while \[ \frac{d}{dx} f(x)^2 = 2f(x) f'(x) \]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8yes, abs is not differntiable at 0.dw:1434682783559:dw and I guess my notation suffices. I would denote everything differently for just my self.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean, it's clear that: \[ \frac{d}{dx}f(x)^2 \neq \left(\frac{d}{dx}f(x)^2\right) f'(x) \]When \(f'(x) \neq 1\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe I should make a derivative notation guide, I've been thinking about it for a while.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8dw:1434682963657:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.8k i got to go now. I will add more stuff. NAd hopefully my notation would be more clear:) Tnx for your input.
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