find the derivative of this!?

Question

y= ln ((x+1)(2x+9))

Answer 1

(ln u)'= u'/u

[ln (uv)]' = (uv)'/(uv) = (uv' + u'v)/(uv) = uv'/uv + u'v/uv = v'/v + u'/u

Or:

ln (uv) = ln u + ln v

[ln (uv)]' = (ln u + ln v)' = (ln u)' + (ln v)' = u'/u + v'/v

Ilusion

Answer 2

y= ln ((x+1)(2x+9))
=> y = ln (x + 1) + ln (2x + 9)
=> dy/dx = 1/(x + 1) + 2/(2x + 9)

Answer 3

We know that d(ln u)/du = 1/u Now let u = (x+1)(2x+9)
then by multiplying we get u = 2x²+11x+9
and du/dx = 4x+11 and if y = ln u, then dy/du = 1/u
By the chain rule, dy/dx = (dy/du)(du/dx)
so since dy/du = 1/u, or 1/(2x²+11x+9)
dy/dx = 1/(2x²+11x+9) times (4x11)
then dy/dx = (4x+11)/(2x²+11x+9)
Hope this helps.

Answer 4

(2(x + 1) + (2x + 9))/((x+1)(2x + 9))
(4x + 11)/(2x² + 11x + 9)

Doug