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Mathematics 8 Online
OpenStudy (anonymous):

Need help taking this derivative, will write the equation below

OpenStudy (anonymous):

\[\sqrt{\left| x \right|}+x/10\] I understand how to take this derivative when x is positive, but what about when x is negative?

OpenStudy (anonymous):

\[\frac{ d }{ dx }\left| f \left( x \right) \right|=\frac{ f \left( x \right) }{ \left| f \left( x \right) \right| }f \prime \left( x \right)\]

OpenStudy (anonymous):

Hmm, idk what to do with that.

OpenStudy (anonymous):

\[\frac{ d }{ dx}\sqrt{\left| x \right|}=\frac{ d }{ dx }\left( \left| x \right| \right)^{\frac{ 1 }{ 2 }}=\frac{ 1 }{ 2}\left( \left| x \right| \right) ^{\frac{ -1 }{ 2 } }\frac{ d }{ dx }\left| x \right|\]

OpenStudy (anonymous):

can you solve further ?

OpenStudy (anonymous):

Possibly, so is \[\left| x \right| = \sqrt{x ^{2}}\] ? And I just solve it this way?

OpenStudy (anonymous):

\[\frac{ d }{ dx }\left(\sqrt{x^2} \right)=\frac{ d }{ dx }\left( x^2 \right)^{\frac{ 1 }{ 2 }}=\frac{ 1 }{ 2 }\left( x^2 \right)^{\frac{ -1 }{ 2 }}2 x=\frac{ x }{ \left( x^2 \right) ^{\frac{ 1 }{ 2 }} }=\frac{ x }{ \left| x \right| }\]

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