Objetivos

OBJETIVO # 6

DERIVADAS TRIGONOMÉTRICAS E INVERSAS

RESOLVER LAS SIGUIENTES DERIVADAS:

y=sec^2⁡2x dy/dx=4 sec^2⁡2x tan⁡2x

y=ln⁡√(sec⁡〖4x 〗 ) dy/dx=2 tan⁡4x

y=csc^3⁡(2x-3) dy/dx=-6 csc^3⁡(2x-3) cot⁡(2x-3)

〖y=ln〗⁡cot⁡(x/a) dy/dx=(-1)/(a sin⁡〖x/a〗 cos⁡〖x/a〗 )

y=sec⁡2x/x dy/dx=(sec⁡2x (2x tan⁡2x-1))/x^2

y=ln⁡cot⁡(x/2) dy/dx=(-1)/(2 sin⁡〖x/2〗 cos⁡〖x/2〗 )

y=e^(x^2 ) cos⁡2x dy/dx=2e^(x^2 ) (x cos⁡2x-sin⁡2x )

y=ln⁡√(tan⁡x ) dy/dx=sec^2⁡x/(2 tan⁡x )

y=√((1+cos⁡x)/(1-cos⁡x )) dy/dx=1/(cos⁡x-1)

y=e^x sin⁡x dy/dx=e^x (cos⁡x+sin⁡x )

y=√x arc cot⁡〖x/4〗 dy/dx=(arc cot⁡〖x/4〗)/(2√x)-(4√x)/(16+x^2 )

y=x^2 arc csc⁡√x dy/dx=2x arc csc⁡√x-x/(2√(x-1))

y=x arc sin⁡x dy/dx=e^x arc cos⁡x-e^x/√(1-x^2 )

y=√(arc sin⁡2x ) dy/dx=1/√((1-4x^2 )arc sin⁡2x )

y=arc sin⁡(cos⁡2x ) dy/dx=

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