## SolomonZelman one year ago uncommon or complicated derivatives

1. SolomonZelman

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

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

I will be adding more stuff later on.

4. anonymous

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

where is the square root ?

6. SolomonZelman

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

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

8. anonymous

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

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

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

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

12. SolomonZelman

lol

13. SolomonZelman

|dw:1434682963657:dw|

14. SolomonZelman

k i got to go now. I will add more stuff. NAd hopefully my notation would be more clear:) Tnx for your input.