# f(x) = (x+2)*ln x finf f'(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

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**

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