Question

Calculus Challenge Problem: Chain Rule Derivatives

Answer 1

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

Answer 2

Medal and fan for first.

Answer 3

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
\[f'(x)=\frac{ \frac{ 1+\frac{ 1 }{ 2\sqrt{x} } }{ 2\sqrt{x+\sqrt{x}} }+1 }{ 2\sqrt{x+\sqrt{x+\sqrt{x}}} }\]

Answer 4

Nice, you got the right steps and answer.

Sorry, someone is talking to me in physics.
That is one I am a bit late.