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

2 Answers

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neela | High School Teacher | (Level 3) Valedictorian

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


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)}.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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)