Find dy/dx if f (x) = (x + 1) ^2x.

u = 2x + 3

du = 2

v = 3x + 2

dv = 3

y = u/v

y' = (v * du - u * dv) / v^2

y' = ((3x + 2) * 2 - (2x + 3) * 3) / (3x + 2) ^2

y' = (6x + 4 - 6x - 9) / (3x + 2) ^2

y' = - 5 / (3x + 2) ^2

(2x + 3) / (3x + 2) = >

(1/3) * (6x + 9) / (3x + 2) = >

(1/3) * (6x + 4 + 5) / (3x + 2) = >

(1/3) * (6x + 4) / (3x + 2) + (1/3) * 5 / (3x + 2) = >

(1/3) * 2 * (3x + 2) / (3x + 2) + 5 / (3 * (3x + 2)) = >

2/3 + 5 / (9x + 6)

u = 9x + 6

u' = 9

2 / 3 + 5 * u^ (-1)

Derive

0 + 5 * (-1) * u^ (-2) * u'

-5 * 9 / (9x + 6) ^2 = >

-5 * 9 / (9 * (3x + 2) ^2) = >

-5 / (3x + 2) ^2

Im in middle school to btw