tolong bantu nomor 14 yaaa
Matematika
Mitha1211
Pertanyaan
tolong bantu nomor 14 yaaa
1 Jawaban
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1. Jawaban hendrisyafa
[tex]y= (\frac{sinx}{cosx}. \frac{1}{cosx}) ^{3}[/tex]
[tex]y= \frac{sin^{3}x}{cos^{6}x}.[/tex]
u = sin³x
u' = 3 sin²x cos x
v = cos⁶x
v' = 6 cos⁵x. (-sin x)
= -6 sin x cos⁵x
[tex]y'= \frac{u'v-v'u}{v^{2}} [/tex]
[tex]y'= \frac{(3 sin^{2}x cosx) (cos^{6}x)-(-6 sin x cos^{5}x)(sin^{3}x)}{(cos^{6}x)^{2}} [/tex]
[tex]y'= \frac{3 sin^{2}x cos^{7}x +6 sin^{4}x cos^{5}x}{cos^{12}x} [/tex]
[tex]y'= \frac{cos^{5}x(3 sin^{2}x cos^{2}x +6 sin^{4}x)}{cos^{12}x} [/tex]
[tex] y'=\frac{3 sin^{2}x cos^{2}x +6 sin^{4}x}{cos^{7}x} [/tex]
[tex] y'=\frac{3 sin^{2}x cos^{2}x }{cos^{7}x} + \frac{6 sin^{4}x}{cos^{7}x} [/tex]
[tex] y'=\frac{3 sin^{2}x }{cos^{5}x} + \frac{6 sin^{4}x}{cos^{7}x} [/tex]
[tex] y'= \frac{3sin^{2}x}{cos^{2}x} (\frac{1}{cos^{3}x} + \frac{2 sin^{2}x}{cos^{5}x}) [/tex]
[tex] y'= 3tgn^{2}x(sec^{3}x + 2 tgn^{2}x sec^{3}x)[/tex]
[tex] y'= 3tgn^{2}x sec^{3}x + 6 tgn^{4}x sec^{3}x[/tex]