Question

Derivar las siguientes funciones
y=Sen³x2
y=tg 6x
y=Sec(8x+1)
y=(Sen x) (Cos 2x)
y=Cos2x/Sen6x
SON DERIVADAS TRIGONOMÉTRICAS
AYUDA PORFAVORRR!
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Answer 1

Explicación paso a paso:

1.

$y' = (3sen^{2} x)(cosx)(2)+(0)(sen^{3}x)\\y' = 6senx^{2} cosx$

2.

$y' = sec^{2}(6x)(6)\\ y' =6sec^{2}(6x)$

3.

$y' = tan(8x+1)sec(8x+1)(8)\\y' =8tan(8x+1)sec(8x+1)\\y'= 8\frac{sin(8x+1)}{cos(8x+1)}\frac{1}{cos(8x+1)}\\ y' =\frac{8sin(8x+1)}{cos^{2}(8x+1) }$

4.

$y'=cosx(cos(2x))+(-sen(2x))(2)(senx)\\y'=cosxcos2x-2sen2xsenx$

5.

$y' = \frac{(-sen(2x))(2)(sen(6x))-(cos(6x)(6)(cos(2x))}{(sen(6x))^{2} }\\y' = \frac{-2sen(2x)sen(6x)-6cos(6x)cos(2x)}{(sen(6x))^{2}}$