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anonymous
 one year ago
Calculus Challenge Problem: Chain Rule Derivatives
anonymous
 one year ago
Calculus Challenge Problem: Chain Rule Derivatives

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}\] Find f'(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Medal and fan for first.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first we imagine \[\sqrt{x+\sqrt{x}}=m\] \[f(x)=\sqrt{x+m}\] \[f'(x)=\frac{ 1+m' }{ 2\sqrt{x+\sqrt{x+\sqrt{x}}} }\] then we know that \[m'=\frac{ 1+\frac{ 1 }{ 2\sqrt{x} } }{ 2\sqrt{x+\sqrt{x}} }\] so \[f'(x)=\frac{ \frac{ 1+\frac{ 1 }{ 2\sqrt{x} } }{ 2\sqrt{x+\sqrt{x}} }+1 }{ 2\sqrt{x+\sqrt{x+\sqrt{x}}} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nice, you got the right steps and answer. Sorry, someone is talking to me in physics. That is one I am a bit late.
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