# If f(x) = ln(x)/(x^2), find f'(1).

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`f(x)=lnx/x^2`

`f'(x)=1/(x^2) d/dxlnx+ln(x) *(-2)x^-3`

`f'(x)=1/x^3-(2lnx)/x^3`

`f'(x)=(1-2lnx)/x^3`

`f'(1)=(1-2ln1)/1^3`

As ln (1)=0

`f'(1)=1`

`f(x) = lnx/(x^2)`

`f'(x) = (1/x^3) - ((2*lnx)/(x^3))`

`or, f'(x) = (1/x^3)*(1 - 2*lnx)`

`Now, f'(1) = (1/1)*(1-2*ln1)`

`or, f'(1) = 1`

**Note:- ln1 = 0**