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.
