tolong ya d jawab secepatnya ini PR turunan dari : 1)y=tan x^2 2)y=tan^2x 3)y=cottan (1-2x^2)

1.
y = tan(x^2)
Aturan rantai:
[tex]y'=\tan'(x^2).(x^2)' \\ y'=\sec^2(x^2).2x \\ y'=2x\sec^2(x^2)[/tex]

2.
y = tan^2 x

[tex]y'=(\tan^2(x))' \\ y'=(\frac{\sin^2(x)}{\cos^2(x)})' \\ y'=\frac{\cos^2(x)(2\sin(x)\cos(x))-\sin^2(x)(2\cos(x)\sin(x))}{\cos^4(x)}[/tex]
[tex]y'=\frac{-2\cos^3(x)\sin(x)-2\cos(x)\sin^3(x)}{\cos^4(x)} \\ y'=\frac{-2\cos(x)\sin(x)(\sin^2(x)+\cos^2(x))}{\cos^4(x)} \\ y'=\frac{-2\sin(x)}{\cos^3(x)} = \frac{-2\sin(x)}{\cos(x)}\times \frac{1}{\cos^2(x)} \\ y'=-2\tan(x).\sec^2(x)[/tex]

3.
[tex]y=\cot(1-2x^2) \\ $Misal : $1-2x^2=u \\ y=\cot(u)=\frac{\cos(u)}{\sin(u)} \\ y'=\frac{-\sin^2(u).u'-\cos^2(u).u'}{\sin^2(u)}=-\frac{u'(\sin^2(u)+\cos^2(u))}{\sin^2(u)} \\ y'=-\frac{u'}{\sin^2(u)}=-\csc(u).u' \\ y'=-\csc(1-2x^2).-4x=4x\csc(1-2x^2)[/tex]

koreksi nomer 2 dan 3. seharusnya gini :

2) y = tan²x = (tan(x))²
dg menggunakan aturan rantai, turunan pertamanya adalah :
y ' = 2tan(x) (sec²x) = 2tan(x)sec²(x)
(remember : derivative of tan is sec²)

3) y = cotan(1 - 2x²)
y ' = -4x (-cosec²(1-2x²)) = 4x cosec²(1-2x²)
(remember : derivative of cotan is -cosec²)