What is the derivative of ln[(sqrt(x+4)*(x+3)^2)/(sqrt3(x+5)) + 50?

We  use  the following  formula , rules and
notations::

d/dxf(x) = f'(x) or
(f(x))'

d/dx (kx^n) =
k*nx^(n-1).

d/dx{ f(x)*g(x) } =  f'(x)g(x) -
f(x)g('x).

d/dx{ f(x)/g(x)} = [f'(x)*g(x)-
f(x)*g'(x)}/(g(x))^2.

d/dx f(g(x)) =  (df/dg )g'(x) , also
called chain rule

d/dx ln f(x) = {1/f(x)
}f'(x)

..............

Given
 f(x) =  ln[(sqrt(x+4)*(x+3)^2)/(sqrt3(x+5)) + 50?

RHS : 50
is treated as the 2nd term. And sqrt3(x+3) is treated as
(sqrt3)(x+5).

LHS : the one sided rectangular bracket [ is
ignored as there is no right pair inthe expression.

WE
consider:

f(x)  = u(x)/v(x)
+50,

where u(x) =
ln(sqrt(x+4)*(x+3)^2).

Therefore f'(x) = {u(x)/v(x)}'
+(50)'

f'(x) = {u'(x)v(x)-u(x)v,(x)}/{v(x)}^2
..............(1), as (50)' = 0.

u'(x) =
1/sqrt(x+4)*(x+3)^2 * {sqrt(x+4)*(x+3)^2}'

u'(x) =
{(1/2)(x+4)^(-1/2)*(x+3)^2 +
(x+4)^(1/2)*2(x+3)}/sqrt(x+4)*(x+3)^2

u'(x)={(x+3)^2
+4(x+4)(x+3)}/2(x+4)(x+3)^2

u'(x) = {x^2+6x+9+4x^2+28x+48}/
2(x+4)(x+3)62

u'(x) =
(5x^2+34x+57)/2(x+4)(x+3)^2

u'(x) =
(5x+19)(x+3)/2(x+4)(x+3) ^2

u'(x)
=(5x+19)/2(x+4)...........(2)

v(x) =
(sqrt3)(x+5)

v'(x) =
(sqrt3)(x+5)'

v'(x) =
sqrt3.

Substituting in (1) , the values
obtained in (2) and (3), we get:

f'(x) =
{{(5x+19)/2(x+4)}*(sqrt3)(x+5) + ln(sqrt(x+4)*(x+3)^2) *(sqrt3)(x+5)}/
((sqrt3)(x+5))^2

f'(x) = {sqrt3(5x+19)(x+5)
+2(sqrt3)(x+4)ln(sqrt(x+4)*(x+3)^2)}/3(x+5)^2.