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

1. anonymous

$f(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}$ Find f'(x)

2. anonymous

Medal and fan for first.

3. anonymous

first 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 $f'(x)=\frac{ \frac{ 1+\frac{ 1 }{ 2\sqrt{x} } }{ 2\sqrt{x+\sqrt{x}} }+1 }{ 2\sqrt{x+\sqrt{x+\sqrt{x}}} }$

4. anonymous

Nice, you got the right steps and answer. Sorry, someone is talking to me in physics. That is one I am a bit late.