Question

please help
find the derivative of

Answer 1

\[y=\sqrt{2x+\cos^2(2x)}\]

Answer 2

i really dont know how to do this one

Answer 3

@hartnn

Answer 4

\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 5

\[y'=\frac{ 1 }{ 2 }(2x+\cos ^{2}(2x))^{-\frac{ 1 }{ 2 }}(2+2\cos(2x)(-2\sin(2x))\]

Answer 6

i this example \(f(x)=2x+\cos^2(2x)\)

Answer 7

and \(f'(x)=2-2\cos(2x)\sin(2x)\) by the chain rule three times

Answer 8

Feynman's looks correct. I believe sat73 forgot the 2 from the argument.

Answer 9

thank u everyone

Answer 10

Lol @nervegrinder my name is not really Feynman, Feynman was a legendary physicist and those words you said would have been typical of Feynman when he was alive.

Answer 11

Feynman's is correct, but you could simplify to:
\[ \frac{ dy}{ dx }=\frac{ \cos^2(2x) }{ \sqrt{2+\cos^2(2x)} }\]
I have a feeling you can simplify this even further, maybe to one trigonometric function, but I am unable to do so.

Answer 12

Isaiah, I know who Feynman is. I was just referencing responses trying to clarify who gave the correct answer.