# `f(x)= (2sqrt(x))ln(x)` Find `f'(x)` and `f'(2)` .

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`f(x) = (2sqrtx) lnx`

To solve for f'(x), let's use the product formula of derivatives which is `(u*v)' = uv' + u'v ` .

So let,

`u=2sqrtx` and v`=lnx`

`u=2x^(1/2)`

And their corresponding derivatives are:

`u'= 2*1/2x^(-1/2) ` and `v'=1/x`

`u'=x^(-1/2)`

Substitute u, v, u' and v' to the product formula of derivatives.

`f'(x) = 2x^(1/2)*1/x + x^(-1/2) lnx`

`f'(x)=(2x^(1/2))/x + (lnx)/x^(1/2)`

`f'(x)= 2/x^(1/2) + (lnx)/x^(1/2)`

`f'(x)= (2+lnx)/x^(1/2)`

So we have,

`f'(x)= (2+lnx)/sqrtx`

To determine f '(2), substitute x=2 to f '(x) .

`f'(2)=(2+ln2)/sqrt2`

`f'(2)=(sqrt2(2+ln2))/2`

`f'(2)= (2sqrt2+sqrt2ln2)/2`

`f'(2)=sqrt2+ (sqrt2ln2)/2`

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**Hence, `f'(x) = (2+lnx)/sqrtx` and `f'(2)=sqrt2 + (sqrt2ln2)/2` . **

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