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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The derivative of a composed function has to follow the chain rule:

f(u(v(x))) = f'(u)*u'(v)*v'(x)

We notice that the given function is a sum of 2 functions: f + g:

ln[(sqrt(x+4)*(x+3)^2)/(sqrt3(x+5)] + 50

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

g(x) = 50

We'll apply the property of derivative of a sum:

the derivative of the sum is the sum of derivatives:

(f+g)' = f' + g'

{ln[(sqrt(x+4)*(x+3)^2)/(sqrt3(x+5)] + 50}' = {ln[(sqrt(x+4)*(x+3)^2)/(sqrt3(x+5)]}' + (50)'

Now, for the function f, we'll note sqrt(x+4)*(x+3)^2/(sqrt3(x+5) by u(x).

u(x) = sqrt(x+4)*(x+3)^2/(sqrt3(x+5)

f(u(x)) = ln u

f'(u(x)) = (ln u)'

(ln u)' = u'/u

u'(x) = [sqrt(x+4)*(x+3)^2/sqrt3(x+5)]'

First, we'll calculate the derivative of numerator:

[sqrt(x+4)*(x+3)^2]' = [sqrt(x+4)]'*(x+3)^2+sqrt(x+4)*[(x+3)^2]'

[sqrt(x+4)*(x+3)^2]' = (x+3)^2/2sqrt(x+4)+2(x+3)*sqrt(x+4)

Now, we'll calculate the derivative of the ratio:

[sqrt(x+4)*(x+3)^2/(sqrt3(x+5)]' = [sqrt(x+4)*(x+3)^2]'*sqrt3(x+5) - sqrt(x+4)*(x+3)^2*[sqrt3(x+5)]'}/3(x+5)

[sqrt(x+4)*(x+3)^2/(sqrt3(x+5)]' ={{(x+3)^2/2sqrt(x+4)+2(x+3)*sqrt(x+4)}*sqrt3(x+5) - sqrt(x+4)*(x+3)^2/2sqrt3(x+5)]}/3(x+5)

u' = {{(x+3)^2/2sqrt(x+4)+2(x+3)*sqrt(x+4)}*sqrt3(x+5) - sqrt(x+4)*(x+3)^2/2sqrt3(x+5)]}/3(x+5)

[ln[(sqrt(x+4)*(x+3)^2)/(sqrt3(x+5)] + 50]' = u'(x)/u(x)

neela | High School Teacher | (Level 3) Valedictorian

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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.