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f(x) = (x+2)*ln x   finf f'(1)

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seedrasoso | Student, Grade 10 | eNoter

Posted August 6, 2010 at 8:59 AM via web

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f(x) = (x+2)*ln x   finf f'(1)

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neela | High School Teacher | Valedictorian

Posted August 6, 2010 at 12:35 PM (Answer #1)

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f(x) = (x+2)*lnx,

To find f'(1).

Solution:

We use

f(u(x)*V(x)) = u(x)v'(x)+u'(x)v(x) ,

(x^n +k)' = (x^n)' = nx^(n-1)  and (lnx)' = 1/x .

Therefore,

f'(x) = {(x+2)*lnx}' = (x+2)(lnx)' +(x+2)'lnx

= (x+2)+1/x

= x+2+1/x

= (x^2+2x+1)/x

=(x+1)^2/x 

f'(1) = (1+1)^2/1 = 2^2/1 = 4

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giorgiana1976 | College Teacher | Valedictorian

Posted August 6, 2010 at 8:06 PM (Answer #2)

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To calculate the value of the derivative of the function, for x = 1, we'll have to differentiate the function.

To differentiate the function, we'll use the rule of the product:

f'(x) = (x+2)'*ln x  + (x+2)*(ln x)'

f'(x) = ln x + (x+2)/x

Now, we can substitute the variable by the value 1.

f'(1) =  ln 1 + (1+2)/1

We know, by definition, that ln 1 = 0.

f'(1) = 0 + 3/1

f'(1) = 3

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 6, 2010 at 9:01 AM (Answer #3)

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f(x) = (x+2) * ln x

Let f(x) = u*v

u= (x+2) ==> u' = 1

v= ln x ==> v' = 1/x

f'(x) = u'v + uv'

        = 1*lnx + (x+2)*(1/x)

        = ln x + 1 + 2/x

f'(1) = ln 1 + 1 + 2/1

        = 0 + 1 + 2

        = 3

==> f'(1) = 3

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