Question

uncommon or complicated derivatives

Answer 1

At first, the definition of the absolute value is: \(\large\color{slate}{ \displaystyle \left|~B~\right| =\sqrt{B~^2~} }\)

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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:

\(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~\left|f(x) \right|~~\right) =\frac{2f(x) }{2\left|f(x) \right| } \times f'(x) }\)

\(\large\color{slate}{ \displaystyle \frac{d}{dx} \left(~~\left|f(x) \right|~\right) =\frac{f(x) }{\left|f(x) \right| } \times f'(x) }\)

Answer 2

So, you can take home the following formula.

\(\large\color{slate}{ \displaystyle \frac{d}{dx} \left|~f(x)~\right| =\frac{f(x) }{\left|f(x) \right| } \times f'(x) }\)

Answer 3

I will be adding more stuff later on.

Answer 4

You should have wrote:\[
\frac{d}{dx} \bigg( f(x)^2 \bigg) = \frac{d}{d(f(x))} \bigg( f(x)^2 \bigg) f'(x)
\]

Answer 5

where is the square root ?

Answer 6

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"

Answer 7

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

Answer 8

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)

\]

Answer 9

yes, abs is not differntiable at 0.
and I guess my notation suffices. I would denote everything differently for just my self.

Answer 10

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

Answer 11

Maybe I should make a derivative notation guide, I've been thinking about it for a while.

Answer 12

lol