Differentiate the function 2(1+sqrtx)-ln(1+sqrtx)^2

giorgiana1976 | Student

Since we recognize composed functions in the given expression, 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) = 2(1+sqrtx) - ln(1+sqrtx)^2

We'll consider the function ln(1+sqrtx)^2 as a composed function an we'll differentiate it using the chain rule:

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

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

We'll simplify and we'll get:

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

We'll remove the brackets:

[ln(1+sqrtx)^2]' = 1/(x + sqrt x)

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

f'(x) = [2(1+sqrtx) - ln(1+sqrtx)^2]'

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

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

We'll eliminate like terms from numerator:

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

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

We'll simplify and we'll get:

f'(x) = 1/(1+sqrtx)

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