# Differentiate y = (3-square root x)+ln(3-square root x)^2

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To differentiate y = (3-square root x)+ln(3-square root x)^2.

Solution:

dy/dx = d/dx {3-sqrtx)+d/dx{ln(3-sqrtx)^2}....(1).

d/dx(3-sqrtx) = d/dx(-sqrtx) = -1/(2sqrt x)....(2)

d/dx ln{3-sqrtx)^2 = 2*ln(3-sqrtx)* d/dx(3- sqrt x).

d/dxln (3-sqrtx)62 = 2ln(3-sqrtx)*{-1/(2 sqrtx }...(3)

We add (2) and (3): d/dx (y) = {-1/(2sqrtx)} {1+2ln(3-sqrtx)}.

Therefore (d/dx )y = {-1/(2sqrtx)}{1+2ln (3-sqrtx)}.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The given expression is made of composed functions and we'll differentiate the expression using the chain rule.

If we have the composed function u(v(x)), then the chain rule is:

[u(v(x))]' = u'(v)*v'(x)

f(x) = (3-sqrtx) + ln(3-sqrtx)^2

We'll consider the function ln(3-sqrtx)^2 as a composed function an we'll differentiate it using the chain rule:

[ln(3-sqrtx)^2]' = [1/(3-sqrtx)^2]*[(3-sqrtx)^2]'*(3-sqrtx)'

[ln(3-sqrtx)^2]' = [2(3-sqrtx)/(3-sqrtx)^2]*(-1/2sqrtx)

We'll simplify and we'll get:

[ln(3-sqrtx)^2]' = -1/sqrt x*(3-sqrtx)

We'll remove the brackets:

[ln(3-sqrtx)^2]' = -1/(3sqrtx - x)

[ln(3-sqrtx)^2]' = 1/(x - 3sqrtx)

Now, we'll substitute the result in the expression of the derivative of f(x):

f'(x) = [(3-sqrtx) + ln(3-sqrtx)^2]'

f'(x) = -1/sqrtx - 1/sqrt x*(3-sqrtx)

f'(x) = (-3+sqrt x - 1)/sqrt x*(3-sqrtx)

We'll combine like terms from numerator:

f'(x) = (sqrt x - 4)/sqrt x*(3-sqrtx)

f'(x) = sqrt x/sqrt x*(3-sqrtx) - 4/sqrt x*(3-sqrtx)

We'll simplify and we'll get:

f'(x) = 1/(3-sqrtx) - 4/sqrt x*(3-sqrtx)