derivatives

[Flic]

last problem for the night, i promise

deriving the following

\(\displaystyle f(x) = (sec2x)^3 tan2x\)

\(\displaystyle 3(sec2x)^2 (2sec2xtan2x)(tan2x) + (sec2x)^3 2(sec2x)^2\)

\(\displaystyle 6(sec2x)^3 (tan2x)^2 + 2(sec2x)^5\)

\(\displaystyle 2(sec2x)^3 (3(tan2x)^2) + (sec2x)^2)\)

\(\displaystyle 2(sec2x)^3 (3 \frac{(sin2x)^2 +1 }{(cos2x)^2})\)

what next? :?

 

[Unco]

G'day again, Flic.

You have successfully differentiated, so now the goal is to play around and get it as 'simple' as possible?

. If we take off from your second-to-last step:

. . \(\displaystyle \L f'(x) = 2\sec^3{2x} (3\underbrace{tan^2{2x}} + \sec^2{2x})\)

. . . . . . . And use the identity
. . . . . . . . . . . .. \(\displaystyle \L \tan^2{2x} = \sec^2{2x} - 1\)

. . . . . We have
. . . . . . . . \(\displaystyle \L = \, 2sec^3{2x}(3\sec^2{2x} - 3 + \sec^2{2x})\)

. . . . . . . . \(\displaystyle \L = \, \, 2sec^3{2x}(4\sec^2{2x} - 3)\)
. . . . . . . . \(\displaystyle \L = \, \, \, \, 8sec^5{2x} - 6sec^3{2x}\)

 

[soroban]

Hello, Flic!

I'd say you did great . . .

Differentiate the following:

\(\displaystyle f(x)\:=\\sec^{^3}2x)(\tan 2x)\)

\(\displaystyle f'(x)\:=\:3\cdot\sec^{^2}2x\cdot(2\cdot\sec 2x\cdot\tan 2x)(\tan2x)\,+\,\sec^{^3}2x\cdot(2\cdot\sec^{^2}2x)\)

\(\displaystyle f'(x)\:=\:6\cdot\sec^{^3}2x\cdot\tan^{^2}2x\,+\,2\cdot\sec^{^5}2x\)

\(\displaystyle f'(x)\:=\:2\cdot\sec^{^3}2x\cdot [3\cdot\tan^{^2}2x\,+\,\sec^{^2}2x]\)

At this point, I wouldn't change to sines and cosines; nothing will cancel.

We could replace \(\displaystyle \tan^{^2}2x\) with \(\displaystyle \sec^{^2}2x\,-\,1\)

. . and get: \(\displaystyle \:2\cdot\sec^{^3}2x\cdot[4\cdot\sec^{^2}2x\,-\,3]\;\) . . . but that's it.