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SolomonZelman

  • one year ago

uncommon or complicated derivatives

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  1. SolomonZelman
    • one year ago
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    At 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: \(\large\color{slate}{ \displaystyle \left|~B~\right| =\sqrt{B~^2~} }\) \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\left|f(x) \right|~\right) =\frac{1}{2\left|f(x) \right| } \times ~\left[2\times f(x) ~\right] \times f'(x) }\) now algebra: \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\left|f(x) \right|~~\right) =\frac{2f(x) }{2\left|f(x) \right| } \times f'(x) }\) \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~~\left|f(x) \right|~\right) =\frac{f(x) }{\left|f(x) \right| } \times f'(x) }\)

  2. SolomonZelman
    • one year ago
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    So, you can take home the following formula. \(\large\color{slate}{ \displaystyle \frac{d}{dx} \left|~f(x)~\right| =\frac{f(x) }{\left|f(x) \right| } \times f'(x) }\)

  3. SolomonZelman
    • one year ago
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    I will be adding more stuff later on.

  4. anonymous
    • one year ago
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    You should have wrote:\[ \frac{d}{dx} \bigg( f(x)^2 \bigg) = \frac{d}{d(f(x))} \bigg( f(x)^2 \bigg) f'(x) \]

  5. SolomonZelman
    • one year ago
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    where is the square root ?

  6. SolomonZelman
    • one year ago
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    d(f(x)).... I thought it would be more understandable to read d/dx ( ...) to be like "oh I am differentiating with respect to x"

  7. anonymous
    • one year ago
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    \[ \begin{split} \frac{d}{dx}\left(\begin{cases} x &x>0\\ -x &x<0 \end{cases} \right) &= \begin{cases} 1 &x>0\\ -1&x<1 \end{cases} \\ &=\begin{cases} \text{sgn}(x) &x \neq 0\\ \text{undef} \end{cases} \\ &=\frac{|x|}{x} &= \frac{x}{|x|} \end{split} \]Once you have this, chain rule can be used for \(|f(x)|\).

  8. anonymous
    • one year ago
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    The problem is that \[ \frac{d}{d(f(x))} f(x)^2 = 2f(x) \]while \[ \frac{d}{dx} f(x)^2 = 2f(x) f'(x) \]

  9. SolomonZelman
    • one year ago
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    yes, abs is not differntiable at 0.|dw:1434682783559:dw| and I guess my notation suffices. I would denote everything differently for just my self.

  10. anonymous
    • one year ago
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    I mean, it's clear that: \[ \frac{d}{dx}f(x)^2 \neq \left(\frac{d}{dx}f(x)^2\right) f'(x) \]When \(f'(x) \neq 1\).

  11. anonymous
    • one year ago
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    Maybe I should make a derivative notation guide, I've been thinking about it for a while.

  12. SolomonZelman
    • one year ago
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    lol

  13. SolomonZelman
    • one year ago
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    |dw:1434682963657:dw|

  14. SolomonZelman
    • one year ago
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    k 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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