tolong dibantu ya kk pake caranya
Jawab:
a.[tex]y' = \frac{-2}{(x-3)^{2} }[/tex]
b. [tex]y' = 2x^{2} + 7x+6[/tex]
Penjelasan dengan langkah-langkah:
a. [tex]y= \frac{x-1}{x-3}[/tex] jika [tex]y = \frac{u}{v}[/tex] maka [tex]y=\frac{u'. v -u . v'}{v^{2} }[/tex]
[tex]u' = 1\\v' = 1[/tex]
[tex]y' = \frac{1 (x-3) - 1(x-1)}{(x-3)^{2} }[/tex]
[tex]y' = \frac{x-3 - x +1}{(x-3)^{2} }[/tex]
[tex]y' = \frac{-2}{(x-3)^{2} }[/tex]
b. [tex]y =( x^{2} + 3x-5) (x+2)[/tex]
[tex]y' =( 2x+3) (x+2)[/tex]
[tex]y' = 2x^{2} + 4x+3x+6[/tex]
[tex]y' = 2x^{2} + 7x+6[/tex]
[answer.2.content]
