taking derivative of absolute values?

[sarahjohnson]

Hi can anyone explain to me the general method of taking the derivative of absolute values and getting 2 answers when x is positive and when it is negative. I looked online but didn't quite understand. Thank You!


y=f(x)=

|x|+

x

7

that's the square root of absolute value of x, plus x/7

 

[mathmari]

When $\displaystyle x>0$ , $\displaystyle \left | x \right | = x$ and when $\displaystyle x<0$ , $\displaystyle \left | x \right | = -x$ .
So you have to take these both cases.

 

[HallsofIvy]

This is $\displaystyle f(x)= \sqrt{|x|}+ \frac{x}{7}$. As long as x> 0 that is the same as $\displaystyle f(x)= x^{1/2}+ \frac{x}{7}$, for x< 0 it is $\displaystyle f(x)= (-x)^{1/2}+ \frac{x}{7}$. The function is NOT DIFFERENTIABLE at x= 0.

 

[DrPhil]

Hi can anyone explain to me the general method of taking the derivative of absolute values and getting 2 answers when x is positive and when it is negative. I looked online but didn't quite understand. Thank You!


y=f(x)=

|x|+

x

7

that's the square root of absolute value of x, plus x/7

Perhaps you will recognize the two cases if you change from absolute value notation to a piecewise function:

\(\displaystyle \displaystyle f(x) = \begin{cases} \sqrt{-x}+x/7 & \text{ if }x < 0 \\
\sqrt{x}+x/7 & \text{ if }x \ge 0\end{cases}\)

 

[sarahjohnson]

ok that makes sense! thanks everyone

 

[lookagain]

Hi can anyone explain to me the general method of taking the derivative of absolute values . . .

sarahjohnson, I would prefer that you stay with the information with the other posts in this thread for your understanding.


However, if you want a more compact-looking derivative, then you can look at my work, but it is relatively complicated.



Let k be any real number.


Some rules:


1) $\displaystyle \ \ |k| \ = \ \sqrt{k^2}$

2) $\displaystyle \ \ \sqrt{k^2} \ \ne \ k$

3) $\displaystyle \ \ \sqrt{k^2} \ = \ (k^2)^{\frac{1}{2}}$

4) $\displaystyle \ \ (\sqrt{k \ })(\sqrt{k \ }) \ = \ k$

5) $\displaystyle \ \ ( \ |k| \ )( \ |k| \ ) \ = \ k^2$



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f(x) = $\displaystyle \ \sqrt{|x| \ }$



f(x) = $\displaystyle \ \sqrt{\sqrt{x^2 \ } \ }$



f(x) = $\displaystyle \ \sqrt{\bigg(x^2\bigg)^{\frac{1}{2}} \ }$



f(x) = $\displaystyle \ \bigg[\bigg(x^2\bigg)^{\frac{1}{2}}\bigg]^{\frac{1}{2}}$



I am going to use the power rule and the chain rule for derivatives:



$\displaystyle f \ '(x) \ = \ \dfrac{1}{2}\bigg[\bigg(x^2\bigg)^{\frac{1}{2}}\bigg]^{\frac{-1}{2}}*\bigg[\dfrac{1}{2}\bigg(x^2\bigg)^{\frac{-1}{2}}\bigg]\bigg(2x\bigg)$



$\displaystyle f \ '(x) \ = \ \dfrac{x}{2\bigg[\bigg(x^2\bigg)^{\frac{1}{2}}\bigg]^{\frac{1}{2}}*\bigg(x^2\bigg)^{\frac{1}{2}}}$


$\displaystyle f \ '(x) \ = \ \dfrac{x}{2\bigg(|x|\bigg)^{\frac{1}{2}}*|x|}$




$\displaystyle f \ '(x) \ = \ \dfrac{x}{2\sqrt{|x| \ }*|x|}$



$\displaystyle f \ '(x) \ = \ \dfrac{x}{2\sqrt{|x| \ }*|x|} \cdot \dfrac{\sqrt{|x| \ }}{\sqrt{|x| \ }}$



$\displaystyle f \ '(x) \ = \ \dfrac{ \ x\sqrt{|x| \ }}{2( \ |x|*|x| \ )}$



$\displaystyle f \ '(x) \ = \ \dfrac{ \ x\sqrt{|x| \ }}{2x^2}$



$\displaystyle \boxed{ \ f \ '(x) \ = \ \dfrac{ \ \sqrt{|x| \ }}{2x} \ }$$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$And, as was pointed out elsewhere, the function is not differentiable at x = 0.