Question

HELP SOMEBODY!!
derivate this!!

Answer 1

\[2\cos ^{3} (\sqrt{e ^{4t}})\]

Answer 2

you could change the sqrt to an exponent .. but its seems to be just a practice in chain rule application

Answer 3

say

\[f=2cos^3(u)\]

what is the derivative of f with respect to u?

Answer 4

\[\frac{df}{du}=[2cos^3(u)]'\]
\[\frac{df}{du}\frac{du}{dt}=[2cos^3(u)]'\frac{du}{dt}\]
\[\frac{df}{dt}=[2cos^3(u)]'\frac{du}{dt}\]
\[\frac{df}{dt}=[2cos^3(u)]'~u'\]

Answer 5

But How should I derivate e\[\sqrt{x ^{4t}}\]

Answer 6

thats \[\sqrt{e^{4t}}\]
or simply
\[\sqrt{e^{4t}}\color{red}{\implies}e^{4t/2}\color{red}{\implies}e^{2t}\]

Answer 7

if u = e^(2t); u' = 2 e^(2t)

Answer 8

\[\frac{df}{dt}=[2cos^3(u)]'~u'\]
\[\frac{df}{dt}=[2cos^3(\sqrt{e^{4t}})]'~(\sqrt{e^{4t}})'\]
\[\frac{df}{dt}=[2cos^3(\sqrt{e^{4t}})]'~(e^{2t})'\]
\[\frac{df}{dt}=-12e^{2t}~~\sin(\sqrt{e^{4t}})\cos^2(\sqrt{e^{4t}})\]